Question

A ball is dropped from the roof of a tower of height h .
The total distance covered by it in last seconed of its motion is equal to the didtsnce covered by it in first three seconds,what is the value of h?

Answer 1

I'm not sure but this maybe the solution....
Let, the distance covered by the ball in last second =x
Therefore the velocity during the last second or final velocity(v)= x/1 m/s.=x.
Initial velocity(u) when ball is dropped is =0.
Given, the ball covers x distance in first 3 seconds.

using equation : s=u×t + 1/2×a×t^2
x=0 + 1/2×9.8×(3)^2
x=44.1m
Therefore, v=44.1 m/s

using equation: v^2=u^2 +2×a×s
44.1×44.1=0+2×9.8×h.
h=(44.1)^2/19.6
h=99.225m(ANS)

THE VALUE OF acceleration due to gravity(g) is taken as 9.8 m/s^2 , so answer may vary if g is taken as 10 m/s^2, then answer will be 101.25m.