Cheenta
How 9 Cheenta students ranked in top 100 in ISI and CMI Entrances?
Learn More

A Bouncing Ball

Try this problem, useful for Physics Olympiad, based on a bouncing ball.

The Problem:

A ball is dropped from a height (h) above a horizontal concrete surface. The coefficient of restitution for the collision involved is (e). What is the time after which the ball stops bouncing?

Discussion:

The time required for the free fall of the ball is (\sqrt{\frac{2h}{g}}). Then the time taken for rise and next fall will be (2\sqrt{\frac{2h}{g}}e).
Time taken for one more rise and fall will be (2\sqrt{\frac{2h}{g}(e^2)}). Thus, the total time for which the ball will be in motion will be
$$ \sqrt{\frac{2h}{g}}+2\sqrt{\frac{2h}{g}e(1+e+e^2+....)}$$ $$=\sqrt{\frac{2h}{g}(1+\frac{2e}{1+e})}$$ $$=\sqrt{\frac{2h}{g}(\frac{1-e}{1+e})}$$

Knowledge Partner

Cheenta is a knowledge partner of Aditya Birla Education Academy
Cheenta

Cheenta Academy

Aditya Birla Education Academy

Aditya Birla Education Academy

Cheenta. Passion for Mathematics

Advanced Mathematical Science. Taught by olympians, researchers and true masters of the subject.
JOIN TRIAL
support@cheenta.com