A ball is dropped on to floor from a height of 5m. It rebounds to a height of 1.25m. If the ball is in contact with floor for 0.01 secs, what is the average acceleration during contact?


When the ball is dropped from a height of 5m, it gains gravitational potential energy. This energy is converted into kinetic energy as the ball falls towards the ground. When the ball hits the ground, it experiences a force that brings it to rest. This force is the normal force exerted by the ground on the ball.

When the ball rebounds, it loses some of its kinetic energy due to the collision with the ground. This loss of energy causes the ball to rebound to a height that is less than its original height.

The time of contact between the ball and ground is very small, so we can assume that the acceleration of the ball during this time is constant.


To find the average acceleration of the ball during contact with the ground, we can use the following formula:

a = (v_f - v_i) / t

Where:
a = acceleration
v_f = final velocity
v_i = initial velocity
t = time

Initially, the ball is at rest, so its initial velocity is zero. When the ball hits the ground, it rebounds with a velocity that is equal in magnitude but opposite in direction to its initial velocity.

Therefore, the final velocity of the ball can be calculated as follows:

v_f = -1 * v_i

The time of contact between the ball and ground is given as 0.01 seconds.

Substituting these values into the formula, we get:

a = (-1 * 0 - 1.25) / 0.01

a = -125 m/s^2

Since the acceleration is negative, it means that the direction of acceleration is opposite to the direction of motion. This indicates that the ball is decelerating during contact with the ground.


The average acceleration of the ball during contact with the ground is -125 m/s^2. This means that the ball is decelerating during contact.

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