Question

A ball is dropped from a height h. It rebounds from the ground a number of times. If coefficient of resistution is e, to what height does it go after nth rebounding?

Please answer urgently.

Answer 1

Answer:

(a)e2nh(b)hen(c)he2n(d)h/e2n

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A)  

For an object bouncing off a stationary object, such as a floor, coefficient of restitution, e=h′h−−−√

⇒ The height to which it bounces h′=e2h

Similarly, after it rebounds for the nth time, the height to which it rises h′=e2nh

Answer: e2nh

 

A)  

e=velocity of separation/velocity of approach

so,  in this case as ground is in rest so,

e=v/u only

every time it collides final velocity becomes initial thus power of 'e' increases.

for n collisions: final velocity(v)=(e^n)(initial velocity)

initial velocity=(2gh)^1/2

so, (v)=(e^n)((2gh)^1/2

thus, as it is uniform motion so,

2gH=v^2

2gH=(e^2n)2gh

thus,   H=(e^2n)h

so, option (a) is correct!

Explanation: