A body is dropped from a height h above the ground. Find the ratio of distances fallen in first one second, firt two seconds, first three seconds, also find the ratio of distance fallen in `1^(st)` second, in `2^(nd)` second, in `3^(rd)` second etc.

A body is dropped from a height h above the ground. Find the ratio of

1.	A body is dropped from a height h above the ground. Find the ratio of distances fallen in first one second, firt two seconds, first three seconds, also find the ratio of distance fallen in `1^(st)` second, in `2^(nd)` second, in `3^(rd)` second etc.
Answer» From second equation of motion, i.e h = `(1)/(2) "gt"^(2)" "(h= ut + (1)/(2)"gt"^(2) and u=0)` `h_1 : h_2 : h_3 ...... = (1)/(2) g(1)^(2) : (1)/(2)g(3)^(2) = 1^(2) : 2^(2) : 3^(2) .......... = 1 : 4 : 9 : ....... . ` Now from th expression of distance travelled in `n^(th)` second `S_n = u + (1)/(2)a (2n-1)` here `u =0, a= g" "` So `" "S_n= (1)/(2)g(2n-1)` therefore `S_1 : S_2 : S_3 ......... = (1)/(2) g ( 2xx1- 1) : (1)/(2)g (2xx 2 -1) : (1)/(2)g( 2xx 3 -1)= 1 : 3: 5 .............. `

Discussion

No Comment Found

Related InterviewSolutions

The position of a particle moving on X-axis is given by `x =At^(2) + Bt + C` The numerical values of A, B and C are 7, -2 and 5 respectively and SI units are used. Find (a) The velocity of the particle at t= 5 (b) The acceleration of the particle at t =5 (c ) The average velocity during the interval t = 0 to t = 5 (d) The average acceleration during the interval t = 0 to t = 5
A particle goes from point A to point B, moving in a semicircle of radius 1 m in 1 second. Find the magnitude of its average velocity.
A particle is dropped from the top of a tower. During its motion it covers `(9)/(25)` part of height of tower in the last 1 second Then find the height of tower.
The distance travelled by a particle in time t is given by `s = (2.5 t^(2))` m. Find (a) the average speed of the partion during time 0 to 0.5 s and (b) the instantaneous speed at t = 5.0s.
A body is dropped from a height h above the ground. Find the ratio of distances fallen in first one second, firt two seconds, first three seconds, also find the ratio of distance fallen in `1^(st)` second, in `2^(nd)` second, in `3^(rd)` second etc.
The velocity of a particle is given by `v=(2t^(2)-4t+3)m//s` where t is time in seconds. Find its acceleration at t=2 second.
A particle moves on straight line according to the velocity-time graph shown in fig. Calculate - (i) Total distance covered (ii) Average speed (iii) In which part of the graph the acceleration is maximum and also find its value. (iv) Retardation
Straight distance between a hotel and a railway stations is 10 km, but circular route is followed by a taxi coven 23 km in 28 minute. What is average speed and magnitude of average velocity ? Are they equal ?
For a particle moving with constant acceleration, prove that the displacement in the `n^(th)` second is given by `s_(n^(th)) = u + (a)/(2)(2n-1)`
A cricketer can throw a ball to a maximum horizontal distance of 100 m. How high above the ground can the cricketer throw the ball, with the same speed ?

A body is dropped from a height h above the ground. Find the ratio of distances fallen in first one second, firt two seconds, first three seconds, also find the ratio of distance fallen in `1^(st)` second, in `2^(nd)` second, in `3^(rd)` second etc.

Discussion

No Comment Found

Related InterviewSolutions

Reply to Comment