Question

A cricket ball is thrown upward with a velocity of $60\mathrm{m}{\mathrm{s}}^{-1}$ ($\mathrm{a}=-10\mathrm{m}{\mathrm{s}}^{-2}$ ).

Calculate the maximum height reached by the ball.

Answer 1

Step 1:- Given data

The initial velocity of the ball $\mathrm{u}=60\mathrm{m}{\mathrm{s}}^{-1}$

Acceleration of ball $a=-10\mathrm{m}{\mathrm{s}}^{-2}$

At the maximum height final velocity $\mathrm{v}=0\mathrm{m}{\mathrm{s}}^{-1}$

Step 2:- Formula used

We want to calculate the maximum height attains by the ball

So we apply the third equation of motion

${\mathrm{v}}^2={\mathrm{u}}^2+2\mathrm{as}$

Step 3:- Calculation For the maximum height

$$\begin{gathered} {\mathrm{v}}^2={\mathrm{u}}^2+2\mathrm{as} \\ {(0\mathrm{m}{\mathrm{s}}^{-1})}^2={(60\mathrm{m}{\mathrm{s}}^{-1})}^2+2\times (-10)(\mathrm{m}{\mathrm{s}}^{-2})\times \mathrm{s} \\ 0=3600-2\times 10\times \mathrm{s} \\ 2\times 10\times \mathrm{s}=3600 \\ \mathrm{s}=180\mathrm{m} \end{gathered}$$

hence The maximum height reached by the ball is 180 m.