Question

A ball thrown vertically upwards from the ground falls back to ground in $3s$ . Determine the velocity of projection of ball and maximum height attained by it.

Answer 1

Step 1: Given data

The time of flight of the ball is $3s$

The velocity of the ball at a maximum height, $v=0m{s}^{-1}$.

Take the initial velocity as $um{s}^{-1}$.

(Note: Consider the downward direction as negative throughout the motion of the ball and the acceleration due to the gravity of the earth as $g=9.8m{s}^{-1}$)

Step 2: Find the initial velocity of projection of the ball

Time to reach the maximum height, $t=\frac{timeofflight}{2}=\frac{3}{2}s$

As we know that the first equation of motion is, $v=u+at$

On substituting all the given data to the above equation we get,

$\Rightarrow v=u-gt$

$\Rightarrow 0=u-9.8\times \frac{3}{2}$

$\Rightarrow u=14.7m{s}^{-1}$

Step 3: Determine the maximum height attained by the ball.

As we know the second equation of motion under gravity is $S=ut+\frac{1}{2}a{t}^2$

On putting the given data in the above equation we get,

$\Rightarrow h=ut-\frac{1}{2}g{t}^2$

$\Rightarrow h=14.7\times \frac{3}{2}-\frac{1}{2}\times 9.8\times {\left(\frac{3}{2}\right)}^2$

$\Rightarrow h=22.050-11.025$

$\Rightarrow h=11.025m$

Thus, the maximum height attained by it is $\mathbf{11}\mathbf{.}\mathbf{025}\mathbf{}\mathit{m}$.