Question

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

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