A ball is projected from the point O with velocity 20 m/s at an angle of $60 °$ with horizontal as shown in figure. At highest point of its trajectory it strikes a smooth plane of inclination $30 °$ 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 $k ?$
$(Take g = 10 m / s^{2})$

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Detailed Solution

Speed of ball before collision is

$v = ucos 60 ° = 20 \times \frac{1}{2} = 10 m / s$

Since, collision is perfectly inelastic (e=0), so the ball will not bounce. It will move along the plane with velocity

$v^{'} = vcos 30 ° = \frac{10 \sqrt{3}}{2} = 5 \sqrt{3} m / s$

∴ Maximum height attained by the ball

$H = \frac{u^{2} \sin^{2} 60 °}{2 g} + \frac{v^{' 2}}{2 g}$

$H = \frac{(20)^{2} {(\frac{\sqrt{3}}{2})}^{2}}{20} + \frac{(5 \sqrt{3})^{2}}{20}$

$= \frac{300}{20} + \frac{75}{20} = \frac{75}{4} m$

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