Question

A body projected vertically up has displacement of 16 m in the first n seconds while it was moving up.its magnitude of displacement in the last n second while falling down is?​

Answer 1

Answer: 16m

Explanation: From second equation of motion,

s = ut + 1/2 at^2

Given:-

s = 16m

time = n

So,

16=un-(1/2)gn^2…......(1). Here, u is initial velocity.

From first equation of motion, v = u + at

Since final velocity at top is 0,

u - gt = 0 ( minus because acceleration against the motion of object)

u/g = t

Therefore, the total time to reach to the top is u/g. The same will be the time for return journey.

t-n = u/g-n

So, velocity at time (u/g-n) is [(u/g)-n]g=(u-ng).

Therefore, distance in last n second is

s=(u-ng)n+(1/2)n^2

 =un-gn^2+(1/2)gn^2

 =un-(1/2)gn^2

 =16 m  (from eq.(1).)