Question

A ball is projected from the ground at an angled of ${45}^o$ with the horizontal surface. It reaches a maximum height of $120m$ and returns to the ground. Upon hitting the ground for the first time, it loses half of its kinetic energy. Immediately after the bounce, the velocity of the ball makes an angle of ${30}^o$ with the horizontal surface. The maximum height it reaches after the bounce, (in metres), is

Answer 1

Given,
Initial maximum hight obtained by the ball= $120m$.
By using projectile motion equation (maximum hight)
$H_1=\frac{u^2{sin}^{2}45}{2g}=120$
$\frac{u^2}{4g}=120$ …. (i)
When half of kinetic energy is lost v = $\frac{u}{\sqrt{2}}$

so again , $H_2=\frac{(\frac{u}{\sqrt{2}}{)}^2{sin}^{2}30}{2g}=\frac{u^2}{16g}$...… (ii)

From (i) and (ii)
$H_2=\frac{1}{4}\left[\frac{u^2}{4g}\right]$, (where, $\frac{u^2}{4g}=120$ from eq. (1))
so, $\frac{1}{4}\times \left(120\right)=30m$

Hence, the maximum hight after the collision will be $30m.$