Question

A ball is thrown vertically upwards from the top of a tower with an initial velocity of $19.6m{s}^{-1}$. The ball reaches the ground after $5s$. Calculate the height of the tower. Take $g=9.8m{s}^{-2}$.

Answer 1

Let us consider point of projection as reference point.
Given, initial velocity, $u=19.6m{s}^{-1}$
$S=-H,\text{Time}\left(t\right)=5s$
$a=-g=-9.8m{s}^{-2}$
$S=ut+\frac{1}{2}a{t}^2$
$-H=19.6\times 5+\frac{1}{2}(-9.8)(5{)}^2$
$-H=98-4.9\times 25$
$-H=98-122.5$
$H=24.5m$
Hence, the height of the tower is equal to $24.5m$.