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 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 $\frac{75}{k}$ meter. Find the value of k? (g = 10 $m/s^2$) ___

Answer 1

$H=\frac{{20}^{2}si{n}^2{60}^{\circ }}{2g}=\frac{60}{4}$

$\frac{1}{2}m(5\sqrt{3}{)}^2=m(\omega )h\Rightarrow h=\frac{15}{4}$
Max Height $=\frac{60}{4}+\frac{15}{4}=\frac{75}{4}$
$\Rightarrow k=4$