A ball (initially at rest) falls vertically for 2s and hits a...

Let the ball strike the plane at A. We assign a co-ordinate system with origin at A. The velocity of the ball while striking the 30^{o} plane = gt = 2g& m/s. The components of u along and perpendicular to the plane are u \sin 30^{o} and u \cos 30^{o} respectively. After impact, the velocity component u \sin 30^{o} along the plane remains unaffected while the component u \cos 30^{o} perpendicular to the plane becomes e(u \cos 30^{o}) on rebound.

Thus velocity of rebound$= e \text{.} 2 \text{g} \text{.} \dfrac{\sqrt{3}}{2} = \sqrt{3} \text{eg}$



The acceleration perpendicular to the plane= - \text{g} \text\cos 3 {0}^{∘} = - \text{g} \sqrt{3} / 2

If t be the time of striking the plane again, then

$0 = \sqrt{3} \text{egt} - \dfrac{1}{2} \dfrac{\sqrt{3}}{2} \text{g} t^{2}$

⇒ \text{ and } t = 4 e = 4 \times \dfrac{5}{8} = 2 \text{.} 5 \text\sec

The distance \text{AB} = u \text\sin 3 {0}^{ο} t + \dfrac{1}{2} \text{g} \text\sin 3 {0}^{ο} t^{2}

$= \left(2 \text{g}\right) \times 2 \text{.} 5 \times \dfrac{1}{2} + \dfrac{1}{2} \times 1 0 \times \dfrac{1}{2} {\left(\dfrac{5}{2}\right)}^{2}$

$= 2 5 + \dfrac{2 5 0}{1 6} = 4 0 \text{.} 6 3 \text{m}$