Question

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

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

Answer 1

Answer: (A)

Speed of ball before collision is
$v=u\cos 60^{\circ }=20\times \frac{1}{2}=10m/s$
Since, collision is perfectly inelastic $(e=0)$, so the ball will not bounce.
It will move along the plane with velocity

$v^{\prime }=v\cos 30^{\circ }=\frac{10\sqrt{3}}{2}=5\sqrt{3}m/s$
$\therefore$ Maximum height attained by the ball
$H=\frac{u^2{\sin }^{2}60^{\circ }}{2g}+\frac{v^{\prime 2}}{2g}$
$H=\frac{(20{)}^2{\left(\frac{\sqrt{3}}{2}\right)}^2}{20}+\frac{(5\sqrt{3}{)}^2}{20}$
$=\frac{300}{20}+\frac{75}{20}=18.75m$