Question

a tennis ball is dropped on the floor from height of 20 m it rebound to a height of 5 m the ball was in contact with floor for 0.01 sec what is average acceleration during contact​

Answer 1

Taking downward motion as in positive direction.Negtive sign here is used because the ball moves against the gravity in bouncing back.

Answer 2

The average acceleration of the ball during contact is $2968\ m/s^2$

Explanation:

Given that,

Initial position, $x_i=20\ m$

Let u is the initial speed of the ball. It can be calculated as :

$u=\sqrt{2gx_i}$

$u=\sqrt{2\times 9.8\times 20}$

u = 19.79 m/s

Final position, $x_f=5\ m$

Let u is the initial speed of the ball. It can be calculated as :

$v=\sqrt{2gx_f}$

$v=\sqrt{2\times 9.8\times 5}$

v = -9.89 m/s (negative because it rebounds)

Time taken, t = 0.01 s

Let a is the average acceleration during the contact. It can be calculated as :

$a=\dfrac{v-u}{t}$

$a=\dfrac{9.89+19.79}{0.01}$

$a=2968\ m/s^2$

So, the average acceleration during contact​ is $2968\ m/s^2$. Hence, this is the required solution.

Learn more :

Topic : Conservation of energy

https://brainly.in/question/12061968