Answer to Question #209271 in Mechanics | Relativity for cr7

Question

A stone is falling freely from rest and the total distance covered by it in the last second of its motion equals the distance covered by it in the first 3 seconds of its motion. How long does the stone remain in air?

Answer 1

Answer:-

$s =ut + 0.5gt^2$ ........1

In first 3sec, let it covers distance "d"m.

$d=0.5g(3)^2 =4.5g$

This distance is same as distance traveled in last second

So, according to formula,

$Sn=u+0.5a(2n−1)$

$4.5g=0.5g(2n−1)$

Here, u=0 and n is number of seconds for which stone is in air.∴n=5 sec

Substituting in eqn 1,

$\boxed{S= 122.5m}$