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 :$\left(i\right)$ the height of the tower, $\left(ii\right)$ the velocity of ball on reaching the ground. Take $g=9.8m{s}^{-2}$.

• A. $\left(i\right)24.5m$, $\left(ii\right)29.4$ $m$ $s^{-1}$
• B. $\left(i\right)24.5m$, $\left(ii\right)19.4$ $m$ $s^{-1}$
• C. $\left(i\right)24.5m$, $\left(ii\right)28$ $m$ $s^{-1}$
• D. $\left(i\right)25m$, $\left(ii\right)29.4$ $m$ $s^{-1}$

Answer 1

Answer: (A)

$\left(i\right)24.5m$, $\left(ii\right)29.4$ $m$ $s^{-1}$

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

$=19.6\times 5-\frac{1}{2}\times 9.8\times 5^2$

$=-24.5m$

Hence, height of tower $=24.5m$

$v=u+at$

$=19.6-9.8\times 5=-29.4m/s$