Question

A body dropped from the top of a tower falls through 40m during the last 2s of its fall. The height of the tower is ($g=10m/s^2$)

Answer 1

Let us suppose that during travelling 40 m in last 2 sec its initial velocity was u.

then, $ut+\frac{1}{2}g{t}^2=40$ ( for last 2 sec)

$\Rightarrow 2u-19.6=40$

$u=29.8m/s$

Now considering from top to this position, $u=0$ and $v=29.8m/s$

then by $v^2-u^2=2gh$ we get,

${29.8}^2-0^2=2\times 9.8\times h$

$h=45.3m$

So total height of tower = $45.3+40=85.30m$.