48 + Interview Questions in PROJECTILE-MOTION IN GENERAL AWARENESS Page 1 InterviewSolution

1.	A threaded rod with `12 turns//cm` and diameter `1.18 cm` is mounted horizontally. A bar with a threaded hole to match the rod is screwed onto the rod. The bar spins at `216 rev//min`. How long will it take for the bar to move `1.50 cm` along the rod ?
Answer» Correct Answer - `(5 s)`

Discussion

2.	A ballon is ascending at the rate `v = 12 km//h` and is being carried horizontally by the wind at `v_(w) = 20 km//h`. If a ballast bag is dropped from the balloon at the instant h = 50 m, determine the time needed for it to strike the ground. Assume that the bag was released from the ballon with the same velocity as the balloon. Also, find the speed with which the bag strikes the ground?
Answer» Correct Answer - B::C `u_x = 20 km//h = 20 xx 5/18 = 5.6 m//s` `u_y = 12 km//h` `= 12 xx 5/18 = 3.3 m//s` Using `S = ut + 1/2 at^2` in vertical direction, we have `-50 = (3.3) t + 1/2 (-10) t^2` Solving this equaiton we get, `t = 3.55 s` At the time of striking with ground, `v_x = u_x = 5.6 m//s` `v_y = u_y + a_yt` `= (3.3) + (-10) (3.55)` `=32.2 m//s` `:. Speed = (sqrt(32.2)^2 + (5.6)^2)` ` ~~ 32.7 m//s`.

Discussion

3.	A particle is dropped from point P at time t = 0. At the same time another particle is thrown from point O as shown in the figure and it collides with the particle P. Acceleration due to gravity is along the negative y-axis. If the two particles collide 2 s after they start, find the initial velocity `v_0` of the particle which was projected from O. Point O is not necessarily on ground.
Answer» Correct Answer - A::B Let `v_x and v_y` be the components of `v_0` along x and y directions. `(v_x)(2) = 2` `:. v_x = 1 m//s` `v_(y) (2) =10` or `v_(y) = 5 m//s` `v_0 = (sqrt (v_(x)^(2)+ v_(y)^2))` `=(sqrt26) m//s` `tan theta = v_y//v_x = 5//1` `:. theta = tan^(-1) (5)`.

Discussion

4.	Two bodies are thrown with the same intial velocity at angles ` theta ` and `(90^@ - theta)` respectively with the horizontal, then their maximum height are in the ratioA. 1 : 1B. ` sintheta : costheta`C. `sin^2theta:cos^2theta`D. `costheta:sintheta`
Answer» Correct Answer - C `H_(theta) = (u^2sin^2theta)/(2g)` `H_(90-theta) = (u^2sin^2(90-theta))/(2g) = (u^2cos^2theta)/(2g)` `:. H_(theta)/H_(90-theta) = (sin^2theta)/(cos^2theta)` .

Discussion

5.	Identify the correct statement related to the projectile motion.A. It is uniformly accelerated everywhereB. It is uniformly accelerated everywhere except at the highest position where it is moving with constant velocityC. Acceleration is never perpendicular to velocityD. None of the above
Answer» Correct Answer - A a = g = constant for small heights.

Discussion

6.	A particle is projected with velocity u at angle `theta` with horizontal. Find the time when velocity vector is perpendicular to initial velocity vector.
Answer» Correct Answer - A::C Given, `v_\|_u` `rArr v.u=0` `rArr (u+at).u=0`......(i) Substituting the proper values in Eq.(i), we have `[{( ucos theta)hat(i)(usin theta)hat(j)}+(-ghat(j))t].[(u cos theta)hat(i)+(u sin theta)hat(j)]=0` `rArr u^(2)cos^(2)theta+u^(2)sin^(2)theta-(ug sin theta)t=0` `rArr u^(2)(sin^(2) theta+cos^(2) theta)=(ug sin theta)t` Solving this equation, we get `t=(u)/(g sin theta)=(u cosec theta)/(g)`

Discussion

7.	A grass hopper can jump maximum distance `1.6m.` It spends negligible time on ground. How far can it go in `10(sqrt2)`s?A. 45 mB. 30 mC. 20 mD. 40 m
Answer» Correct Answer - D Maximum range is obtained at `45^@`. `u^2/g = 1.6 or u=4m//s` `T = (2usin45^@)/g = (2xx4xx(1//sqrt2))/10` `= 0.4(sqrt2)s` Number of jumps in given time, `n = t/T = (10(sqrt2))/(0.4(sqrt2)) = 25` `:.` Total distance travelled `= 1.6 xx 25 = 40 m.`

Discussion

8.	Two stones are projected with the same speed but making different angles with the horizontal. Their horizontal ranges are equal. The angle of projection of one is `pi/3` and the maximum height reached by it is 102 m. Then the maximum height reached by the other in metres isA. 76B. 84C. 56D. 34
Answer» Correct Answer - D `H_1 = (u^2sin^2theta)/(2g)` `:. 102 = ((u)^2sin^(2)60^@)/20` `:. u = 52.2 m//s` Other stone should be projected at `90^@-theta` or `30^@` form horizontal. `:. H_2 = (u^2 sin^2 30^@)/(2g)` `=((52.2)^2(1//4))/20` =34m.

Discussion

9.	The maximum vertical height of a projectile is 10 m. If the magnitude of the initial velocity is `28 ms^(-1)`, what is the direction of the initial velocity. ? Take `g = 9.8 ms^(-2)`.
Answer» Correct Answer - `(30^(@))`

Discussion

10.	Show that there are two values of time for which a projectile is at the same height. Also show mathematically that the sum of these two times is equal to the time of flight.
Answer» Correct Answer - A For vertically upward motion of a projectile, `y=(u sin alpha)t - 1/2 (g t^2) or 1/2 g t^2 -(usin alpha)t+y=0` This is a quadratic equation in t. Its roots are `t_(1)=(u sing alpha-sqrt(u^(2)sin^(2)alpha-2gy))/(g)` and `t_(2)=(us sin alpha+sqrt(u^(2)sin^(2)alpha-2gy))/(g)` `:. t_(1)+t_(2)=(2u sin alpha)/(g)=T` (time of flight of the projectile).

Discussion

11.	Find time of flight and range of the projectile along the inclined plane as shown in figure. `(g = 10 m//s^2)`
Answer» Correct Answer - A::C `T = (2usin(alpha -beta))/(g cos beta)` `((2)(20sqrt2)sin(45^@ -30^@))/((10)(cos 30^@)) = 1.69s` `R = u^2/(gcos^2beta) [sin(2alpha-beta)-sinbeta]` `=(20sqrt(2))^2/((10)cos^2 30^@) [sin(2xx45^@-30^@)- sin 30^@]` `=39m`.

Discussion

12.	In the figure shown, find. (a) time of flight of the projectile along the inclined plane. (b) range OP.
Answer» Correct Answer - A::B (a) Horizontal component of initial velocity, `u_x = 20sqrt(2) cos 45^@ = 20 m//s` Vertical component of initial velocity, `u_y = 20 sqrt(2) sin 45^@ = 20 m//s` Let the particle, strikes the inclined plane at P after time t, then horizontal displacement `QP = u_x t = 20t` In vertical displacement, `u_yt or 20t ` is upward and `1/2 g t^2 or 5t^2` is downwards. But net vertical displacement is downwards. Hence `5t^2` should be greater than 20 t and therefore, `OQ = 5t^2- 20t` In `DeltaOQP, ` `tan37^@ = OQ/QP` or `3/4 = (5t^2 - 20t)/(20t)` Solving this equation, we get `t= 7s` (b) Range, OP `= (PQ) sec 37^@` `= (20t)(5/4)` Substituting the value of t, we get `OP = 175 m.`

Discussion

13.	In the figure shown, find. (a) time of flight of the projectile along the inclined plane. (b) range OP.
Answer» Correct Answer - A::B (a) Horizontal component of initial velocity, `u_x = 20(sqrt2) cos 45^@ = 20 m//s` Vertical component of initial velocity `u_y = 20(sqrt2) sin 45^@ = 20 m//s` Let the particle strikes at P after time t, then horizontal displacement `OQ = u_xt = 20t ` In vertical displacement, `u_yt` or 20t is upwards and `1/2 g t^2` or `5t^2` is downwards. But net displacement is upwards, therefore 20 t should be greater than `5t^2` and `QP = 20t - 5t^2 ` In `Delta OPQ, ` `tan 37^@ = (QP/OQ)` or, ` 3/4 = ((20t - 5t^2)/(20t))` Solving this equation, we get `t = 1s` (b) Range `OP = OQ sec 37^@` ` =(20t)(5/4)` Substituting t = 1s, we have OP = 25m .

Discussion

14.	Find time of flight and range of the projectile along the inclined plane as shown in figure. `(g = 10 m//s^2)`
Answer» Correct Answer - A::C::D `T = (2usin(alpha +beta))/(g cos beta)` `=(2xx20sqrt(2) sin (45^@ + 30^@))/((10)cos30^@) ~~ 6.31 s` `R = u^2/(g cos^2beta) [sin(2alpha+beta) + sinbeta]` `= (20sqrt(2))^2/((10)cos^2 30^@) [sin(2xx45^@ + 30^@)+ sin30^@]` `=145.71 m`

Discussion

15.	Assertion : If time of flight in a projectile motion is made two times, its maximum height will become four times. Reason : In projectile motion `H prop T^2` , where H is maximum height and T the time of flight.A. (a)If both Assertion and Reason are true and the Reason is correct explanation of the Assertion.B. (b)If both Assertion and Reason are true and the Reason is not the correct explanation of the Assertion.C. (c) If Assertion is true, but the Reason is false.D. (d) If Assertion is false, but the Reason is true.
Answer» Correct Answer - A `T = (2usintheta)/g rArr usintheta = (gT)/2` `H = (u^2sin^2theta)/(2g) = ((gT//2)^2/(2g))` or `H prop T^2`.

Discussion

16.	Assertion : In projectile motion if particle is projected with speed u, then speed of particle at height h would be `(sqrt (u^2 - 2gh)).` Reason : If particle is projected with vertical component of velocity `u_y` . Then vertical component at the height h would be `+- (sqrt (u_(y)^2) - 2gh).`A. (a)If both Assertion and Reason are true and the Reason is correct explanation of the Assertion.B. (b)If both Assertion and Reason are true and the Reason is not the correct explanation of the Assertion.C. (c) If Assertion is true, but the Reason is false.D. (d) If Assertion is false, but the Reason is true.
Answer» Correct Answer - B `v_x = u_x` `v_y = +- (sqrt(u_(y)^2) -2gh)` and `v=(sqrt(v_(x)^2) +v_(y)^2)` and `u^2 = u_(x)^2 + u_(y)^2` .

Discussion

17.	Find the angle of projection for a projectile motion whose rang `R` is (n) times the maximum height `H`.
Answer» Correct Answer - `(theta=tan^(-1).(4)/(n))`

Discussion

18.	A projectile aimed at a mark, which is in the horizontal plane through the point of projection, falls a cm short of it when the elevation is `alpha` and goes b cm far when the elevation is `beta`. Show that, if the speed of projection is same in all the cases the proper elevation is `1/2 sin^(-1) [(bsin2alpha+asin2beta)/(a+b)]` .
Answer» Correct Answer - A `R - a = (u^2 sin 2 alpha)/g` Multiplying with b we have, `bR - ab = (bu^2 sin 2alpha)/g`………..(i) `R + b = (u^2sin2beta)/g` Multiplying with a we have, `aR + ab = (au^2sin 2beta)/g` ……..(ii) Adding Eqs. (i) and (ii) and by putting `R = (u^2sin 2theta)/g,` we get the result.

Discussion

19.	Find the angle of projection of a projectile for which the horizontal range and maximum height are equal.
Answer» Correct Answer - A::D Given, R=H `:. (u^2 sin2alpha)/(g) = (u^2sin^2alpha)/(2g) or 2sin alpha cos alpha = (sin^2alpha)/(2)` or `(sinalpha)/(cosalpha) = 4 or (tanalpha)=4` `:. alpha = tan^-1(4)`

Discussion

20.	A projectile is fired horizontally with velocity of 98 m/s from the top of a hill 490 m high. Find (a) the time taken by the projectile to reach the ground, (b) the distance of the point where the particle hits the ground from foot of the hill and (c) the velocity with which the projectile hits the ground. `(g= 9.8 m//s^2)` .
Answer» Correct Answer - A::B Here, it will be more convenient to choose x and y directions as shown in figure. Here, `u_x = 98 m/s , a_x = 0, u_y =0 `and` a_y = g` (a) At A, `s_y = 490 m`. So, applying `s_y = u_(y)t + 1/2(a_y)t^2` `:. 490 = 0 + 1/2(9.8)t^2 ` `:. t=10s ` (b) ` BA=s_x = u_xt + 1/2(a_x)t^2` or ` BA = (98)(10)+0` or `BA = 980m` (c) `v_x = u_x + a_xt = 98+0 = 98 m//s` `v_y = u_y + a_y t = 0+(9.8)(10)=98m//s` `:. v = (sqrt((v_(x)^2)+(v_(y)^2))) = (sqrt((98)^2 + (98)^2)) = 98 (sqrt2) m//s` and `tan beta = ((v_y)/(v_x)) = 98/98 = 1` `:. beta = 45^@` Thus, the projectile hits the ground with velocity `98(sqrt2) m//s` at an angle of `beta = 45^@` with horizontal as shown in Fig. 7.9.

Discussion

21.	A train is moving on a track at `30ms^-1`. A ball is thrown from it perpendicular to the direction of motion with `30 ms^-1` at `45^@` from horizontal. Find the distance of ball from the point of projection on train to the point where it strikes the ground.A. 90 mB. `90(sqrt3)` mC. 60 mD. `60 (sqrt3)`m
Answer» Correct Answer - A Velocity of train in the direction of train is also `30 m//s`. So there is no relative motion in this direction. In perpendicular direction, `d = R = (u^2sin2theta)/g` `=((30)^2sin90^@)/10 = 90m`.

Discussion

22.	Two particles are thrown from the same point with the same velocity of `49 ms^(-1)`. First is projected making angle `theta` with the horizontal and second at an angle `(90^(@)-theta)`. The second particle is found to rise 22.5 m higher than the first. Find the height to which each particle will rise.
Answer» Correct Answer - `(50 m, 72.5 m)`

Discussion

23.	A heavy particle is projected with a velocity at an angle with the horizontal into the uniform gravitational field. The slope of the trajectory of the particle varies asA. B. C. D.
Answer» Correct Answer - A `y = x tan theta - (gx^2)/(2u^2 cos^2theta)` `:. (dy)/(dx) = (tan theta)-(g/(u^2cos^2theta))x` `:. (dy)/(dx)` versus x graph is a straight line with negative slope and positive intercept.

Discussion

24.	A 500 kg car takes a round turn of radius 50 m with a velocity of 36 km/hr . The centripetal force is
Answer» Correct Answer - `(2 ms^(-2))`

Discussion

25.	A body is projected up such that its position vector varies with time as `r = { 3thati + (4 t - 5t^2)hatj}` m. Here, t is in seconds. Find the time and x-coordinate of particle when its y-coordinate is zero.
Answer» Correct Answer - A::B::C::D y-coordinate of particle is zero when, `4t - 5t^2 = 0` `:. t = 0 and 0.8 s` `x = 3t` at t = 0, x = 0 and at t=0.8 s, x = 2.4 m.

Discussion

26.	A train is moving with a constant speed of `10m//s` in a circle of radius `16/pi`m. The plane of the circle lies in horizontal x-y plane. At time t = 0, train is at point P and moving in counter- clockwise direction. At this instant, a stone is thrown from the train with speed `10m//s` relative to train towards negative x-axis at an angle of `37^@` with vertical z-axis . Find (a) the velocity of particle relative to train at the highest point of its trajectory. (b) the co-ordinates of points on the ground where it finally falls and that of the hightest point of its trajectory. (Take g `= 10 m//s^2, sin 37^@ = 3/5`
Answer» Correct Answer - B::C At t = 0 `v_T = (10hatj)m//s` `v_(ST) = 10 cos 37^@hat k - 10 sin 37^@hati = (8hatk-6hati)m//s` `:. V_s = v_(ST) + v_(T) = (-6hati + 10hatj+8hatk)m//s` (a) At highest point vertical component `(hatk)`of `v_s` will become zero. Hence, velocity of particle at highest point will become `(-6hati+10hatj)m//s`. (b) Time of flight, `T = (2v_Z)/g = (2xx8)/10 = 1.6s` `x = x_i + v_xT` `= 16/pi -6 xx 1.6 = -4.5m` `y = (10)(1.6) = 16 m and z = 0` Therefore , coordinates of particle where it finally lands on the ground are `(-4.5m, 16m, 0)` At highest point, `t = T/2 = 0.8s` `:. x = 16/pi - (6)(0.8) = 0.3m` `y = (10)(0.8)= 8.0m` and `z = (v_(Z)^(2))/(2g) = (8)^2/20 = 3.2m` Therefore , coordinates at highest point are, `(0.3 m, 8.0m, 3.2m)`.

Discussion

27.	A particle is projected at an angle `60^@` with horizontal with a speed `v = 20 m//s`. Taking `g = 10m//s^2`. Find the time after which the speed of the particle remains half of its initial speed.
Answer» Correct Answer - C `v_x = u_x = 20 cos 60^@ = 10 m//s` Given, `v=u/2` `:. 4 v^2 = u^2` or `4(v_(x)^2 + v_(y)^2) = u^2` `:. 4[(10)^2 + v_(y)^2] = (20)^2` or `v_y = 0` Hence, it is the highest point `:. t = T/2 = (usintheta)/g` `=((20)sin60^@)/10` `=(sqrt3) s`.

Discussion

28.	Two particles move in a uniform gravitational field with an acceleration g. At the initial moment the particles were located over a tower at one point and moved with velocities `v_1 = 3m//s and v_2= 4m//s` horizontally in opposite directions. Find the distance between the particles at the moment when their velocity vectors become mutually perpendicular.
Answer» Correct Answer - B `V_1 * V_2 = 0` when` v_1 _\|_ v_2 ` `:. (u_1 + a_1t) * (u_2 + a_2t) = 0` `:. (3hati - 10thatj) * (4hati - 10t hatj) = 0` or `t = (sqrt(0.12)) s` Now in vertical direction they have no relative motion and in horizontal direction their velocities are opposite. `:. d = 3t + 4t = 7t` `=(7) (sqrt(0.12)) m` `~~ 2.5m`

Discussion

29.	Two particles A and B are projected from ground towards each other with speeds `10 m//s` and `5(sqrt2) m//s` at the angle `30^@ and 45^@` with horizontal from two points separated by a distance of 15 m. will they collide or not ?
Answer» For collision to take place, relative velocity of A with respect of B should be along Ab or their vertical components should be same. Vertical components of A `v_1 = 10 sin 30^@ = 5 m//s` and vertical component of B `v_2 = 5(sqrt2) cos 45^@ = 5 m//s` Since, `v_1 = v_2`, so they may collide Now the second condition is , `R_1+R_2 ge d (d=15 cm)` `:. R_1 = ((10)^2sin60^@)/10 = 8.66 m` `R_2 = ((5sqrt(2))^2sin90^@)/10 = 5m` Since, `R_1+R_2 ltd`, so they will not collide.

Discussion

30.	A particle is projected from ground with velocity `20(sqrt2) m//s` at `45^@`. At what time particle is at height 15 m from ground? `(g = 10 m//s^2)`
Answer» Correct Answer - A::C::D Vertical component of initial velocity `u = 20(sqrt2) sin 45^@ = 20 m//s` Now, apply `s = ut + 1/2 at^2` (to find t) in vertical direction, with `s = 15m, u=20 m//s and a = -10 m//s^2`.

Discussion

31.	An aeroplane is flying in a horizontal direction with a velocity `600 km//h` at a height of 1960 m. When it is vertically above the point A on the ground, a body is dropped from it. The body strikes the ground at point B. Calculate the distance AB.
Answer» Correct Answer - `(3333.3 m)`

Discussion

32.	In between two hills of heights 100 m and 92 m respectively. There is a valley of breadth 16 m. If a vehicle jumps from the first hill to the second, what must be its minimum horizontal velocity so that it may not fall into the valley ? Take `g=9 ms^(-2)`
Answer» Correct Answer - `(12 ms^(-1))`

Discussion

33.	Assertion : Projectile motion is called a two dimensional motion, although it takes place in space. Reason : In space it takes place in a plane.A. (a)If both Assertion and Reason are true and the Reason is correct explanation of the Assertion.B. (b)If both Assertion and Reason are true and the Reason is not the correct explanation of the Assertion.C. (c) If Assertion is true, but the Reason is false.D. (d) If Assertion is false, but the Reason is true.
Answer» Correct Answer - A

Discussion

34.	In the above problem, what is the component of its velocity perpendicular to the plane when it strikes at A ?
Answer» `v_y = u_y + a_yt` `=(10 sin30^@) + (-g cos 30^@) T` `= 5 - 10 xx (sqrt3)/2 xx 2/(sqrt3)` `= -5 m//s`.

Discussion

35.	A shell is fired at an angle of `30^(@)` to the horizontal with a velocity of `392 ms^(-1)`. Find the time of flight, horizontal range and maximum height attained.
Answer» Correct Answer - (40 s, 13578.88 m, 1960 m)

Discussion

36.	A particle is projected from the bottom of an inclined plane of inclination `30^@` with velocity of `40 m//s` at an angle of `60^@` with horizontal. Find the speed of the particle when its velocity vector is parallel to the plane. Take `g = 10 m//s^2`.
Answer» Correct Answer - C::D Horizontal component of velocity always remains constant `:. 40 cos 60^@ = v cos 30^@` or `v= 40/(sqrt3) m//s`.

Discussion

37.	A particle is fired horizontally form an inclined plane of inclination `30^@` with horizontal with speed `50 ms^-1`. If `g=10 ms^-2` , the range measured along the incline isA. 500 mB. `1000/3 m`C. `200 (sqrt2)` mD. `100 (sqrt3)` m
Answer» Correct Answer - B `R = (u^2)/(gcos^2beta) [sin(2alpha+beta) + sin beta]` `u = 50m//s, g= 10m//s^2, alpha = 0^@`, `R = (50)^2/(10 cos^2 30^@) [sin(2 xx 0 + 30^@) + sin 30^@]` `= ((2500))/(10 xx (3//4)) (1/2 +1/2)` `=1000/3`.

Discussion

38.	A ball is projecte with a velocity `20 ms^-1` at an angle to the horizontal. In order to have the maximum range. Its velocity at the highest position must beA. `10 ms^-1`B. `14ms^-1`C. `18ms^-1`D. `16ms^-1`
Answer» Correct Answer - B At `45^@` , range is maximum. At highest point, it has only horizontal component of velocity or `20cos45^@ = 14 m//s` .

Discussion

39.	A particle is projected at an angle of elevation `alpha` and after t second it appears to have an elevation of `beta` as seen from the point of projection. Find the initial velocity of projection.
Answer» Correct Answer - A::B::C Let v is the velocity at time t Using `v_y = u_y + a_yt ` `:. v sin beta = usin alpha - g t` ………(i) `v_x = u_x` `:. v cos beta = u cos alpha …..(ii)` From Eqs. (i) and (ii), we get `((ucosalpha)/(cos beta)) sin beta = u sinalpha -g t` On solving we get, `u = (g t cosbeta)/(sin (alpha-beta))`.

Discussion

40.	Two tall buildings are situated 200 m apart. With what speed must a ball be thrown horizontally from the window 540 m above the ground in one building, so that it will enter a window 50 m above the ground in the other ?
Answer» Correct Answer - `(20 ms^(-1))`

Discussion

41.	Assertion : A particle has constant acceleration is x-y plane. But neither of its acceleration components `(a_x and a_y)` is zero. Under this condition particle cannot have parabolic path. Reason : In projectile motion, horizontal component of acceleration is zero.A. (a)If both Assertion and Reason are true and the Reason is correct explanation of the Assertion.B. (b)If both Assertion and Reason are true and the Reason is not the correct explanation of the Assertion.C. (c) If Assertion is true, but the Reason is false.D. (d) If Assertion is false, but the Reason is true.
Answer» Correct Answer - D In a projectile motion along an inclined plane, normally we take x and y-axis along the plane and perpendicular to it. In that case, `a_x and a_y` both are non-zero.

Discussion

42.	The seocnd-hand of a watch is 3.0 cm long. Calculate the linear speed of its top.
Answer» Correct Answer - `(0.314 ms^(-1))`

Discussion

43.	A football is kicked with speed ` 20 ms^(-1)` at a projection angle of ` 45^@` from the ground. A receiver on the goal line `20 m` away in the direction of the kick runs the same instant to meet the ball. What must be his speed, if he has to catch the ball before it hits the ground ? Take ` g= 10 ms^(-2)`.
Answer» Correct Answer - `(5.483 ms^(-1))`

Discussion

44.	A cyclist moves along a circular path of radius 70 m. If he completes one round in 11 s, calculate (i) total length of path, (ii) magnitude of the displacement, (iii) average speed, and (iv) magnitude of average velocity.
Answer» Correct Answer - [(i) 440 m (ii) zero (iii) `40 ms^(-1)` (iv) zero]

Discussion

45.	A football player kicks a ball at an angle of `37^(@)` to the horizontal with an initial speed of `15 ms^(-1)`. Assuming that the ball travels in a vertical plane, calculate (i) the time at which the ball reaches the highest point (ii) the maximum height reached (iii) the horizontal range of the projectile and (iv) the time for which the reached (iii) the horizontal range of the projectile and (iv) the time for which the ball is in air.
Answer» Correct Answer - [(i) 0.92 s (ii) 4.16 m (iii) 21.2 m (iv) 1.84 s]

Discussion

46.	Two particles projected form the same point with same speed u at angles of projection `alpha and beta` strike the horizontal ground at the same point. If `h_1 and h_2` are the maximum heights attained by the projectile, R is the range for both and `t_1 and t_2` are their times of flights, respectively, thenA. `alpha + beta = pi/2`B. `R = 4(sqrt(h_1h_2))`C. `t_1/t_2 = tan alpha`D. `tan alpha = (sqrt (h_1/h_2))`
Answer» Correct Answer - A::B::C::D `alpha + beta = 90^@` or `beta = 90^@ - alpha` `h_1= (u^2sin^2alpha)/(2g) and h_2 = (u^2cos^2alpha)/(2g)` `t_1 = (2usinalpha)/g and t_2 = (2ucosalpha)/g` `R_1 = R_2 = (2u^2sinalphacosalpha)/g = R` .

Discussion

47.	Assertion: A particle follows only a parabolic path if acceleration is constant. Reason : In projectile motion path is parabolic, as acceleration is assumed to be constant at low heights.A. (a)If both Assertion and Reason are true and the Reason is correct explanation of the Assertion.B. (b)If both Assertion and Reason are true and the Reason is not the correct explanation of the Assertion.C. (c) If Assertion is true, but the Reason is false.D. (d) If Assertion is false, but the Reason is true.
Answer» Correct Answer - D In the cases shown below path is straight line, even if a is constant `rarr u` `rarr u` `rarr (a)` `larr (a)`.

Discussion

48.	The angular velocity of a particle moving in a circle of radius `50 cm` is increased in 5 min from `100` revolutions per minute to `400` revolutions per minute. Find the tangential acceleration of the particle.
Answer» Correct Answer - `[(pi)/(30)"rad "s^(-2) (ii) (5pi)/(3)cms^(-2)]`

Discussion

Explore topic-wise InterviewSolutions in .

A threaded rod with `12 turns//cm` and diameter `1.18 cm` is mounted horizontally. A bar with a threaded hole to match the rod is screwed onto the rod. The bar spins at `216 rev//min`. How long will it take for the bar to move `1.50 cm` along the rod ?

Two bodies are thrown with the same intial velocity at angles ` theta ` and `(90^@ - theta)` respectively with the horizontal, then their maximum height are in the ratioA. 1 : 1B. ` sintheta : costheta`C. `sin^2theta:cos^2theta`D. `costheta:sintheta`

Identify the correct statement related to the projectile motion.A. It is uniformly accelerated everywhereB. It is uniformly accelerated everywhere except at the highest position where it is moving with constant velocityC. Acceleration is never perpendicular to velocityD. None of the above

A particle is projected with velocity u at angle `theta` with horizontal. Find the time when velocity vector is perpendicular to initial velocity vector.

A grass hopper can jump maximum distance `1.6m.` It spends negligible time on ground. How far can it go in `10(sqrt2)`s?A. 45 mB. 30 mC. 20 mD. 40 m

Two stones are projected with the same speed but making different angles with the horizontal. Their horizontal ranges are equal. The angle of projection of one is `pi/3` and the maximum height reached by it is 102 m. Then the maximum height reached by the other in metres isA. 76B. 84C. 56D. 34

The maximum vertical height of a projectile is 10 m. If the magnitude of the initial velocity is `28 ms^(-1)`, what is the direction of the initial velocity. ? Take `g = 9.8 ms^(-2)`.

Show that there are two values of time for which a projectile is at the same height. Also show mathematically that the sum of these two times is equal to the time of flight.

Find time of flight and range of the projectile along the inclined plane as shown in figure. `(g = 10 m//s^2)`

In the figure shown, find. (a) time of flight of the projectile along the inclined plane. (b) range OP.

In the figure shown, find. (a) time of flight of the projectile along the inclined plane. (b) range OP.

Find time of flight and range of the projectile along the inclined plane as shown in figure. `(g = 10 m//s^2)`

Find the angle of projection for a projectile motion whose rang `R` is (n) times the maximum height `H`.

Find the angle of projection of a projectile for which the horizontal range and maximum height are equal.

A heavy particle is projected with a velocity at an angle with the horizontal into the uniform gravitational field. The slope of the trajectory of the particle varies asA. B. C. D.

A 500 kg car takes a round turn of radius 50 m with a velocity of 36 km/hr . The centripetal force is

A body is projected up such that its position vector varies with time as `r = { 3thati + (4 t - 5t^2)hatj}` m. Here, t is in seconds. Find the time and x-coordinate of particle when its y-coordinate is zero.

A particle is projected at an angle `60^@` with horizontal with a speed `v = 20 m//s`. Taking `g = 10m//s^2`. Find the time after which the speed of the particle remains half of its initial speed.

Two particles A and B are projected from ground towards each other with speeds `10 m//s` and `5(sqrt2) m//s` at the angle `30^@ and 45^@` with horizontal from two points separated by a distance of 15 m. will they collide or not ?

A particle is projected from ground with velocity `20(sqrt2) m//s` at `45^@`. At what time particle is at height 15 m from ground? `(g = 10 m//s^2)`

An aeroplane is flying in a horizontal direction with a velocity `600 km//h` at a height of 1960 m. When it is vertically above the point A on the ground, a body is dropped from it. The body strikes the ground at point B. Calculate the distance AB.

In between two hills of heights 100 m and 92 m respectively. There is a valley of breadth 16 m. If a vehicle jumps from the first hill to the second, what must be its minimum horizontal velocity so that it may not fall into the valley ? Take `g=9 ms^(-2)`

In the above problem, what is the component of its velocity perpendicular to the plane when it strikes at A ?

A shell is fired at an angle of `30^(@)` to the horizontal with a velocity of `392 ms^(-1)`. Find the time of flight, horizontal range and maximum height attained.

A particle is projected from the bottom of an inclined plane of inclination `30^@` with velocity of `40 m//s` at an angle of `60^@` with horizontal. Find the speed of the particle when its velocity vector is parallel to the plane. Take `g = 10 m//s^2`.

A particle is fired horizontally form an inclined plane of inclination `30^@` with horizontal with speed `50 ms^-1`. If `g=10 ms^-2` , the range measured along the incline isA. 500 mB. `1000/3 m`C. `200 (sqrt2)` mD. `100 (sqrt3)` m

A ball is projecte with a velocity `20 ms^-1` at an angle to the horizontal. In order to have the maximum range. Its velocity at the highest position must beA. `10 ms^-1`B. `14ms^-1`C. `18ms^-1`D. `16ms^-1`

A particle is projected at an angle of elevation `alpha` and after t second it appears to have an elevation of `beta` as seen from the point of projection. Find the initial velocity of projection.

Two tall buildings are situated 200 m apart. With what speed must a ball be thrown horizontally from the window 540 m above the ground in one building, so that it will enter a window 50 m above the ground in the other ?

The seocnd-hand of a watch is 3.0 cm long. Calculate the linear speed of its top.

A cyclist moves along a circular path of radius 70 m. If he completes one round in 11 s, calculate (i) total length of path, (ii) magnitude of the displacement, (iii) average speed, and (iv) magnitude of average velocity.

The angular velocity of a particle moving in a circle of radius `50 cm` is increased in 5 min from `100` revolutions per minute to `400` revolutions per minute. Find the tangential acceleration of the particle.