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

1.	A ball is thrown down wards with velocity V from the top of a tower and it reaches the ground with speed 3V. what is the eight of the tower.(a) V2/g.(b) 2V2/g.(c) 4V2/g. (d) 8V2/g.
Answer» Answer is (c) 4v²/g Use v² - u² = 2ax, here v = 3v, u = v and a = g. hence x = 4v²/g

1.

A ball is thrown down wards with velocity V from the top of a tower and it reaches the ground with speed 3V. what is the eight of the tower.(a) V2/g.(b) 2V2/g.(c) 4V2/g. (d) 8V2/g.

Answer»

Answer is (c) 4v²/g

Use v² - u² = 2ax, here v = 3v, u = v and a = g. hence x = 4v²/g

2.	A ball is thrown horizontally from the top of a tower. What happens to the horizontal component of its velocity.(a) increases (b) decreases(c) remains unchanged(d) first increases then decreases
Answer» Answer is (c) remains unchanged

Discussion

3.	A ball is dropped from the top of a tower the distance covered by it in the last second is (9/25)th of the height of the tower what is the height of the tower.(a) 122.5 m (b) 100.5 m(c) 88.5 m(d) 64.5
Answer» Answer is (a) 122.5 m h = (1/2) gt² and [1 – (9/25)]h = (1/2) g(t - 1)²

3.

A ball is dropped from the top of a tower the distance covered by it in the last second is (9/25)th of the height of the tower what is the height of the tower.(a) 122.5 m (b) 100.5 m(c) 88.5 m(d) 64.5

Answer»

Answer is (a) 122.5 m

h = (1/2) gt² and [1 – (9/25)]h = (1/2) g(t - 1)²

Discussion

4.	The friction of the air causes vertical retardation equal to one tenth of the acceleration due to gravity (take g = 10ms-2) the time of flight will be decreased by(a) 0% (b)1% (c) 9% (d) 11%
Answer» Answer is (c) 9% The time to rise to the top = 91% of u sinθ/g. Also maximum height Hm = 91% of u² sin² θ/2g. And time of fall ≅ using θ/g. Hence time of flight is decreased by 9%.

4.

The friction of the air causes vertical retardation equal to one tenth of the acceleration due to gravity (take g = 10ms-2) the time of flight will be decreased by(a) 0% (b)1% (c) 9% (d) 11%

Answer»

Answer is (c) 9%

The time to rise to the top = 91% of u sinθ/g. Also maximum height Hm = 91% of u² sin² θ/2g. And time of fall ≅ using θ/g.

Hence time of flight is decreased by 9%.

Discussion

5.	Which of the following does not effect the maximum height attained by the projectile ?(a) magnitude of initial velocity(b) acceleration of the projectile(c) angle of the projection(d) mass of the projectile
Answer» Answer is (d) mass of the projectile (d)mass of the projectile

Discussion

6.	The range R of a projectile is same when its maximum heights are h1 & h2. What is the relation between R, h1 & h2?(a) R = √h1 h2 (b) R = √ 2h1 h2 (c) R = 2√h1 h2 (d) R = 4√h1 h2
Answer» Answer is (d) R = 4√h₁ h₂ Range is same for angle of projection θ & 90 - θ R = u² sin2θ/g, h₁ = u² sin² θ/2g h₂= u²cos² θ/2g hence √h₁h₂ = u² sinθ cosθ/2g = 1/4[u² sinθ/g] = R/4

6.

The range R of a projectile is same when its maximum heights are h1 & h2. What is the relation between R, h1 & h2?(a) R = √h1 h2 (b) R = √ 2h1 h2 (c) R = 2√h1 h2 (d) R = 4√h1 h2

Answer»

Answer is (d) R = 4√h₁ h₂

Range is same for angle of projection θ & 90 - θ

R = u² sin2θ/g, h₁ = u² sin² θ/2g

h₂= u²cos² θ/2g

hence √h₁h₂ = u² sinθ cosθ/2g

= 1/4[u² sinθ/g] = R/4

Discussion

7.	A projectile is fired at 30°. with momentum p, neglecting friction, the change in kinetic energy when it returns to the ground will be (a) zero (b) 30% (c) 60% (d) 100%
Answer» Answer is (a) zero on return to the ground the speed is same as on firing. Hence here is no change in kinetic energy.

7.

A projectile is fired at 30°. with momentum p, neglecting friction, the change in kinetic energy when it returns to the ground will be (a) zero (b) 30% (c) 60% (d) 100%

Answer»

Answer is (a) zero

on return to the ground the speed is same as on firing. Hence here is no change in kinetic energy.

Discussion

8.	A projectile is fired with speed u making angle θ with the horizontal. Its potential energy at the highest point is(a) (1/2) mu2 sin2θ(b) (1/2) mu2 cos2θ(c) (1/2) mu2(d) (1/2) mu2 sin2 2θ
Answer» Answer is (a) (1/2) mu² sin²θ The vertical component of velocity is reduced to zero and kinetic energy corresponding to it is converted into potential energy.

8.

A projectile is fired with speed u making angle θ with the horizontal. Its potential energy at the highest point is(a) (1/2) mu2 sin2θ(b) (1/2) mu2 cos2θ(c) (1/2) mu2(d) (1/2) mu2 sin2 2θ

Answer»

Answer is (a) (1/2) mu² sin²θ

The vertical component of velocity is reduced to zero and kinetic energy corresponding to it is converted into potential energy.

Discussion

9.	In the above questions, the change in momentum on return to the ground will be (a) zero (b) 30% (c) 60% (d) 100%
Answer» Answer is (d) 100% ∆p = 2psinθ = 2p sin30° = p. hence ∆p/p = 100%.

9.

In the above questions, the change in momentum on return to the ground will be (a) zero (b) 30% (c) 60% (d) 100%

Answer»

Answer is (d) 100%

∆p = 2psinθ = 2p sin30° = p. hence ∆p/p = 100%.

Discussion

10.	Two balls of same mass are projected one vertically upwards and the other at angle 60°. with the vertical. The ratio of their potential energy at the highest point is (a) 3:2 (b) 2:1 (c) 4:1 (d) 4: 3
Answer» Answer is (c) 4:1 Potential energies at the highest point are equal to the loss in kinetic energies. That is (1/2 mu²) and (1/2) m(u cos60°) = (1/4) x (1/2)mu².

10.

Two balls of same mass are projected one vertically upwards and the other at angle 60°. with the vertical. The ratio of their potential energy at the highest point is (a) 3:2 (b) 2:1 (c) 4:1 (d) 4: 3

Answer»

Answer is (c) 4:1

Potential energies at the highest point are equal to the loss in kinetic energies. That is (1/2 mu²) and (1/2) m(u cos60°) = (1/4) x (1/2)mu².

Discussion

11.	Two balls of same mass are projected one vertically upwards and the other at angle 60°. with the vertical. The ratio of their potential energy at the highest point is In the above question the kinetic energy at the highest point for the second ball is K. what is the kinetic energy for the first ball ?(a) 4K (b) 3K (c) 2K (d) zero
Answer» Answer is (d) zero The velocity of the first ball at the highest point is zero.

11.

Two balls of same mass are projected one vertically upwards and the other at angle 60°. with the vertical. The ratio of their potential energy at the highest point is In the above question the kinetic energy at the highest point for the second ball is K. what is the kinetic energy for the first ball ?(a) 4K (b) 3K (c) 2K (d) zero

Answer»

Answer is (d) zero

The velocity of the first ball at the highest point is zero.

Discussion

12.	A stone thrown upwards with speed u attains maximum height h, another stone thrown upward from the same point with speed 2u attains maximum height H. What is the relation h & H.(a) 2h = H(b) 3h = H (c) 4h = H(d) 5h = H
Answer» Answer is (c) 4h = H Use v² - u² = 2ax. Here 0 - u² = 2gh.. and 0 - 4u² = 2gH.. hence h/H = 1/4.. that is 4h = H.

12.

A stone thrown upwards with speed u attains maximum height h, another stone thrown upward from the same point with speed 2u attains maximum height H. What is the relation h & H.(a) 2h = H(b) 3h = H (c) 4h = H(d) 5h = H

Answer»

Answer is (c) 4h = H

Use v² - u² = 2ax. Here 0 - u² = 2gh.. and 0 - 4u² = 2gH..

hence h/H = 1/4.. that is 4h = H.

Discussion

13.	The velocity of projection of a body is increased by 2%. Others factors remaining unchanged. What will be the percentage change in the maximum height attained.(a) 1%(b) 2%(c) 4%(d) 8%
Answer» Answer is (c) 4% y_m = u² sin² θ/2g. hence ∆y_m/y_m = 2∆u/u, since ∆u/u = 2%, therefore ∆ym/ym = 4.

13.

The velocity of projection of a body is increased by 2%. Others factors remaining unchanged. What will be the percentage change in the maximum height attained.(a) 1%(b) 2%(c) 4%(d) 8%

Answer»

Answer is (c) 4%

y_m = u² sin² θ/2g. hence ∆y_m/y_m = 2∆u/u, since ∆u/u = 2%,

therefore ∆ym/ym = 4.

Discussion

14.	The maximum height attained by the projectile is increased by 5%. Keeping the angle of projection constant. What is the percentage increased in the horizontal range. (a) 5% (b) 10% (c) 15% (d) 20%
Answer» Answer is (a) 5% If y_m be the maximum height, then R = 4y_m cotα.. therefore ∆R/R = ∆ym/ym. Hence percentage increase in range is also equal to 5%.

14.

The maximum height attained by the projectile is increased by 5%. Keeping the angle of projection constant. What is the percentage increased in the horizontal range. (a) 5% (b) 10% (c) 15% (d) 20%

Answer»

Answer is (a) 5%

If y_m be the maximum height, then R = 4y_m cotα.. therefore ∆R/R = ∆ym/ym.

Hence percentage increase in range is also equal to 5%.

Discussion

15.	The range of a projection is maximum. If the range is R, what is the maximum height.(a) 2R (b) R (c) R/2 (d) R/4
Answer» Answer is (d) R/4. Range is maximum for θ = 45°. In such a case y_m = R/4.

15.

The range of a projection is maximum. If the range is R, what is the maximum height.(a) 2R (b) R (c) R/2 (d) R/4

Answer»

Answer is (d) R/4.

Range is maximum for θ = 45°. In such a case y_m = R/4.

Discussion

16.	A projectile is projected with kinetic energy K, it has the maximum possible horizontal range, then its kinetic energy at the highest point will be(a) 0.25 K (b) 0.5 K (c) 0.75 K (d) K.
Answer» Answer is (b) 0.5 K Since range is maximum therefore θ = 45°, hence v_x = ν cos45° = ν/√2. At the highest point the net velocity of the projectile is ν_x.

16.

A projectile is projected with kinetic energy K, it has the maximum possible horizontal range, then its kinetic energy at the highest point will be(a) 0.25 K (b) 0.5 K (c) 0.75 K (d) K.

Answer»

Answer is (b) 0.5 K

Since range is maximum therefore θ = 45°, hence v_x = ν cos45° = ν/√2.

At the highest point the net velocity of the projectile is ν_x.

Discussion

17.	A bullet is fired horizontally with a velocity of 200 ms-1, if the acceleration due to gravity is 10 ms-2, in the first second it fall through a height of(a) 5m (b) 10m (c) 20 m (d) 200 m.
Answer» Answer is (a) 5m Horizontal velocity has no effect on the distance of vertical fall

17.

A bullet is fired horizontally with a velocity of 200 ms-1, if the acceleration due to gravity is 10 ms-2, in the first second it fall through a height of(a) 5m (b) 10m (c) 20 m (d) 200 m.

Answer»

Answer is (a) 5m

Horizontal velocity has no effect on the distance of vertical fall

Discussion

18.	A particle of mass ‘m’ is projected with a velocity ν making angle of 45° with the horizontal. The magnitude of the angular momentum of the projectile about the point of projection when at the maximum height h is(a) zero(b) mν3/4√2g (c) mν3/√2g(d) 2m(2gh3)1/2.
Answer» Answer is (b) mν³/4√2g Velocity at the highest point = ν cos 45° = ν/√2 Maximum height h = ν² sin² 45/2g = ν²/4g. L = angular momentum = [m(ν/√2) x (ν²/4g) = mν³/4√2g Since ν² = 4gh. Therefore L = m(2gh³)^(1/2). Which is not give.

18.

A particle of mass ‘m’ is projected with a velocity ν making angle of 45° with the horizontal. The magnitude of the angular momentum of the projectile about the point of projection when at the maximum height h is(a) zero(b) mν3/4√2g (c) mν3/√2g(d) 2m(2gh3)1/2.

Answer»

Answer is (b) mν³/4√2g

Velocity at the highest point = ν cos 45° = ν/√2

Maximum height h = ν² sin² 45/2g = ν²/4g.

L = angular momentum = [m(ν/√2) x (ν²/4g) = mν³/4√2g

Since ν² = 4gh.

Therefore L = m(2gh³)^(1/2). Which is not give.

Discussion

19.	A gun fires two bullets at 60° & 30° with horizontal. The bullet strikes at same horizontal distance. The ratio of maximum height for the two bullets is in the ratio (a) 2:1(b) 3:1 (c) 4:1 (d) 1:1
Answer» Answer is (b) 3:1 y_m = u² sin² θ/2g and y_m1y_m2 = sin² 60°/sin2 30° = (√3/2)² (1/2)² = 3/1.

19.

A gun fires two bullets at 60° & 30° with horizontal. The bullet strikes at same horizontal distance. The ratio of maximum height for the two bullets is in the ratio (a) 2:1(b) 3:1 (c) 4:1 (d) 1:1

Answer»

Answer is (b) 3:1

y_m = u² sin² θ/2g and y_m1y_m2 = sin² 60°/sin2 30° = (√3/2)² (1/2)² = 3/1.

Discussion

20.	The projectile goes farthest away from the earth, when the angle of projection is (a) 0°. (b) 45°. (c) 90°. (d)180°.
Answer» Answer is (c) 90°. height to which the projectile rises is largest when θ = 90°

20.

The projectile goes farthest away from the earth, when the angle of projection is (a) 0°. (b) 45°. (c) 90°. (d)180°.

Answer»

Answer is (c) 90°.

height to which the projectile rises is largest when θ = 90°

Discussion

21.	Which of the following is the essential characteristic of a projectile?(a) initial velocity inclined to the horizontal.(b) zero velocity at the highest point.(c) constant acceleration perpendicular to the velocity(d) none of the above
Answer» Answer is (d) none of the above

Discussion

22.	A projectile has νx = 20m/s and νy = 10 ms. What will be the angle made by the velocity of the body with the vertical.(a) tan-1 (2.0)(b) tan-1 (1.5)(c) tan-1 (1.0)(d) tan-1 (0.5)
Answer» Answer is (d) tan^-1(0.5) tanθ = νy/νx hence νy =νx.

22.

A projectile has νx = 20m/s and νy = 10 ms. What will be the angle made by the velocity of the body with the vertical.(a) tan-1 (2.0)(b) tan-1 (1.5)(c) tan-1 (1.0)(d) tan-1 (0.5)

Answer»

Answer is (d) tan^-1(0.5)

tanθ = νy/νx

hence νy =νx.

Discussion

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A ball is thrown down wards with velocity V from the top of a tower and it reaches the ground with speed 3V. what is the eight of the tower.(a) V2/g.(b) 2V2/g.(c) 4V2/g. (d) 8V2/g.

A ball is thrown horizontally from the top of a tower. What happens to the horizontal component of its velocity.(a) increases (b) decreases(c) remains unchanged(d) first increases then decreases

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In the above questions, the change in momentum on return to the ground will be (a) zero (b) 30% (c) 60% (d) 100%

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The velocity of projection of a body is increased by 2%. Others factors remaining unchanged. What will be the percentage change in the maximum height attained.(a) 1%(b) 2%(c) 4%(d) 8%

The maximum height attained by the projectile is increased by 5%. Keeping the angle of projection constant. What is the percentage increased in the horizontal range. (a) 5% (b) 10% (c) 15% (d) 20%

The range of a projection is maximum. If the range is R, what is the maximum height.(a) 2R (b) R (c) R/2 (d) R/4

A projectile is projected with kinetic energy K, it has the maximum possible horizontal range, then its kinetic energy at the highest point will be(a) 0.25 K (b) 0.5 K (c) 0.75 K (d) K.

A bullet is fired horizontally with a velocity of 200 ms-1, if the acceleration due to gravity is 10 ms-2, in the first second it fall through a height of(a) 5m (b) 10m (c) 20 m (d) 200 m.

A particle of mass ‘m’ is projected with a velocity ν making angle of 45° with the horizontal. The magnitude of the angular momentum of the projectile about the point of projection when at the maximum height h is(a) zero(b) mν3/4√2g (c) mν3/√2g(d) 2m(2gh3)1/2.

A gun fires two bullets at 60° &amp; 30° with horizontal. The bullet strikes at same horizontal distance. The ratio of maximum height for the two bullets is in the ratio (a) 2:1(b) 3:1 (c) 4:1 (d) 1:1

The projectile goes farthest away from the earth, when the angle of projection is (a) 0°. (b) 45°. (c) 90°. (d)180°.

Which of the following is the essential characteristic of a projectile?(a) initial velocity inclined to the horizontal.(b) zero velocity at the highest point.(c) constant acceleration perpendicular to the velocity(d) none of the above

A projectile has νx = 20m/s and νy = 10 ms. What will be the angle made by the velocity of the body with the vertical.(a) tan-1 (2.0)(b) tan-1 (1.5)(c) tan-1 (1.0)(d) tan-1 (0.5)

The range R of a projectile is same when its maximum heights are h1 & h2. What is the relation between R, h1 & h2?(a) R = √h1 h2 (b) R = √ 2h1 h2 (c) R = 2√h1 h2 (d) R = 4√h1 h2

A stone thrown upwards with speed u attains maximum height h, another stone thrown upward from the same point with speed 2u attains maximum height H. What is the relation h & H.(a) 2h = H(b) 3h = H (c) 4h = H(d) 5h = H

A gun fires two bullets at 60° & 30° with horizontal. The bullet strikes at same horizontal distance. The ratio of maximum height for the two bullets is in the ratio (a) 2:1(b) 3:1 (c) 4:1 (d) 1:1