Question

A ball is projected from the point $O$ with velocity $20m/s$ at an angle of ${60}^o$ with horizontal as shown in the figure. At highest point of its trajectory, it strikes a smooth plane of inclination ${30}^o$ at point $A$. The collision is perfectly inelastic. The maximum height from the ground attained by the ball is

• A. $18.75m$
• B. $15m$
• C. $22.5m$
• D. $20.25m$

Answer 1

Answer: (A)

$18.75m$

Speed of ball before collision is
$v=u\cos {60}^o=20\times \frac{1}{2}=10m/s$
Since, collision is perfectly inelastic $(e=0)$, so the ball will not bounce. It will move along the plane with velocity
$v^{{}^{\prime }}=v\cos {30}^o=\frac{10\sqrt{3}}{2}=5\sqrt{3}m/s$
$\therefore$ Maximum height attained by the ball
$H=\frac{u^2{\sin }^2{60}^o}{2g}+\frac{v^{{{}^{\prime }}^2}}{2g}$
$H=\frac{{\left(20\right)}^2{\left(\frac{\sqrt{3}}{2}\right)}^2}{20}+\frac{{\left(5\sqrt{3}\right)}^2}{20}$
$=\frac{300}{20}+\frac{75}{20}=18.75m$