Question

A ball is projected from a horizontal surface at an angle ${60}^{\circ }$ with horizontal. The maximum height of the projectile is $240\text{m}$ from the ground. Upon hitting the ground for the first time, it loses $75\%$ of its kinetic energy. Immediately after the bounce, the velocity of the ball makes angle of ${45}^{\circ }$ with the horizontal surface. The maximum height of the ball after the bounce is

• A. $60\text{m}$
• B. $120\text{m}$
• C. $240\text{m}$
• D. $180\text{m}$

Answer 1

The correct option is B $120\text{m}$

Given that
Maximum height $\left(H\right)=240\text{m}$
$(K{E}_1{)}_2=0.75\times (K{E}_1{)}_1$
So, from formula
$H=\frac{u^2(\sin {60}^{\circ }{)}^2}{2g}\Rightarrow u^2=\frac{240\times 2g\times 4}{3}=80\times 8g$
Let mass of the ball be $m$.
So, $K{E}_1=\frac{1}{2}m{u}^2=\left(\frac{1}{2}m\right)\times 80\times 8g$
From question, kinetic energy after bounce,
$\left(K{E}_2\right)=0.75\times \left(K{E}_1\right))=\frac{3}{4}\times \frac{1}{2}m\times 80\times 8g=\left(\frac{1}{2}m\right)\times 60\times 8g$

If $\left(K{E}_2\right)=\frac{1}{2}m{u}^{\prime 2}$
$u^{\prime 2}=60\times 8g$
Now, again apply formula for max. height
$H^{\prime }=\frac{(u^{\prime }{)}^2{\sin }^2{\theta }^{{}^{\prime }}}{2g}=\frac{60\times 8g{\left(\sin {45}^{\circ }\right)}^2}{2g}$
$\Rightarrow H^1=60\times 4\times \frac{1}{2}=120\text{m}$