A rubber ball drops from a height 'h'. After rebounding twice from the ground, it rises to h/2. The coefficient of restitution is:
(A) 1/2
(B) (1/2)^(1/2)
(C) (1/2)^(1/4)
(D) (1/2)^(1/6)

To solve this question, we need to use the concept of the coefficient of restitution and the relationship between the rebound heights of the ball after each bounce.
Coefficient of Restitution (e):
The coefficient of restitution is the ratio of the velocity after the rebound to the velocity before the rebound. It is given by:
e=Velocity after bounceVelocity before bouncee = \frac{\text{Velocity after bounce}}{\text{Velocity before bounce}}
When a ball is dropped from a height hh, the velocity just before hitting the ground can be found using the equation for free fall:
v=2ghv = \sqrt{2gh}
where:
• gg is the acceleration due to gravity,
• hh is the height from which the ball is dropped.
First Rebound:
After the first bounce, the ball rises to some height h1h_1. If the coefficient of restitution is ee, the velocity after the first bounce will be ee times the velocity just before impact.
The height to which the ball rises after the first bounce is related to the velocity by the equation:
h1=e2⋅hh_1 = e^2 \cdot h
Second Rebound:
After the second bounce, the ball rises to a height h2h_2, which is given as h2\frac{h}{2}. Therefore, the equation for the second rebound height is:
h2=e2⋅h1=e2⋅e2⋅h=e4⋅hh_2 = e^2 \cdot h_1 = e^2 \cdot e^2 \cdot h = e^4 \cdot h
We are told that h2=h2h_2 = \frac{h}{2}, so:
e4⋅h=h2e^4 \cdot h = \frac{h}{2}
Simplifying:
e4=12e^4 = \frac{1}{2}
Taking the fourth root of both sides:
e=(12)14e = \left( \frac{1}{2} \right)^{\frac{1}{4}}
Conclusion:
The coefficient of restitution ee is:
e=(12)14e = \left( \frac{1}{2} \right)^{\frac{1}{4}}
Thus, the correct answer is:
C) (12)14\left( \frac{1}{2} \right)^{\frac{1}{4}}.