Question

If a particle is dropped from the top of a tower ,its displacement in the first second and in the last second is same. Then average speed of the particle is

Answer 1

A particle is dropped from the top of tower. so, initial velocity of particle, u = 0

displacement in 1st second, s = ut + 1/2at²

= 0 × 1s + 1/2 × 10m/s² × (1s)²

= 5m .......(1)

displacement in last second, s' = $S_t-S_{(t-1)}$

= 1/2 gt² - 1/2 g(t - 1)²

= 1/2g(t² - t² + 2t - 1)

= 1/2 g(2t - 1)

= 1/2 × 10m/s² (2t - 1)

= 5(2t - 1) ......(2)

a/c to question,

displacement in first sec = displacement in last sec

or, 5 = 5(2t - 1)

or, t = 1sec

so, height of tower , s = 1/2 10 × 1² = 5m

so, average speed of particle = total distance/total time taken

= 5m/1s = 5m/s