A girl throws a ball with initial velocity $v$ at an inclination of $45 °$ . The ball strikes a smooth vertical wall at a horizontal distance d from the girl and after rebouncing returns to her hand. The coefficient of restitution between the wall and the ball is $\frac{k g d}{v^{2} - g d}$ where ‘k’ is ____

see full answer

Talk to JEE/NEET 2025 Toppers - Learn What Actually Works!

Real Strategies. Real People. Real Success Stories - Just 1 call away

An Intiative by Sri Chaitanya

answer is 1.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

$T = \frac{d}{(v / \sqrt{2})} + \frac{d}{(e v / \sqrt{2})} = (1 + \frac{1}{e}) \frac{\sqrt{2} d}{v}$

$where T = \frac{2 Vsin 45^{\circ}}{g}$

$\frac{2 v / \sqrt{2}}{g} = (1 + \frac{1}{e}) \frac{\sqrt{2} d}{v} \Rightarrow e = \frac{gd}{v^{2} - gd}$

Watch 3-min video & get full concept clarity

JEE

NEET

Foundation JEE

Foundation NEET

CBSE