Question

A cricket ball is thrown upward with a velocity of 60 m/sec (a = -10 $\mathrm{m}/{\mathrm{s}}^2$ ). Calculate the time taken by the ball to reach the maximum height.

Answer 1

Step 1: Given data

The initial velocity of the ball, $\mathrm{u}=10\mathrm{m}{\mathrm{s}}^{-1}$

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

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

Step 2: Formula used

For the calculating time, we apply the first equation of motion.

The first equation of motion is given as,

$\mathrm{v}=\mathrm{u}+\mathrm{at}$

Step 3: Calculation

From the first equation of motion

$\mathrm{v}=\mathrm{u}+\mathrm{at}$

0 = 60 + (-10)t

0 = 60 - 10t

10t= 60

t =$\frac{60}{10}=6\sec$.

So the ball reaches maximum height after 6 sec.