Question

A perfectly elastic ball is dropped from a high altitude and it acquires a terminal velocity before hitting the ground, from where it bounces off elastically. If air resistance depends on the speed of the ball, its acceleration immediately after the first bounce will be

Answer 1

If Newton's second law were applied to their falling motion, and if a ... But acceleration depends upon two factors: force and mass. ... As such, all objects free fall at the same rate regardless of their mass.

Answer 2

Answer:

Given air resistance is directly proportional to velocity (v)

Therefore, air resistance = kv (where k is a constant)

Initially, the forces on the ball may be defined as :

mg (downwards - due to gravity) - kv (upwards - due to air resistance) = ma

When the ball acquires terminal velocity, acceleration is zero

Therefore, mg - kv' = 0  {where v' = terminal velocity (which is a constant)}

ie, mg=kv'

k = mg/v'   ---> 1

After the bounce of the ball, since the collision is elastic,

FINAL VELOCITY (after bounce of the ball) = INITIAL VELOCITY (before the bounce of the ball) = v'

Now, two forces act on the ball :

(1) Gravity (downwards) = mg

(2) Air Resistance (downwards) = kv'

Therefore,

mg + kv' = ma (where a is the acceleration that we should find)

Substitute k = mg/v'  here

mg + (mg/v')v' = ma

mg + mg = ma

2mg = ma

a = 2g

The answer : acceleration = 2g

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