Question

A stone is released from the top of tower. it covers 24.5 m distance in the last second of its journey. what is the height of tower?

Answer 1

Given

Last second $t=1sec$

Distance covered $S=24.5m$

We know

$S=ut+\frac{1}{2}a{t}^2$

In this case, as the stone is released.

Initial velocity, $u=o$ and

$a=g=9.8m/se{c}^2$

$S=\frac{1}{2}\times 9.8t^2=4.9t^2$

Distance traveled by stone in the last second is

$4.9t^2-4.9(t-1{)}^2=24.5$

$t^2-(t-1{)}^2=5$

$(t-t+1)(t+t-1)=5$

$2t-1=5$

$t=3sec$

Now,

$S=ut+\frac{1}{2}a{t}^2$

$S=0+\frac{1}{2}\times 9.8\times 9$

$S=44.1m$