Question

A tennis ball of mass 60 g is dropped from a height of 2.9 m. It rebounds but loses 23 % of its kinetic energy in the bounce. Ignoring any air resistance. Calculate the velocity at which it leaves the ground.

Answer 1

The velocity at which the ball leave the ground is 6.62 m/s.

Step-by-step explanation:

Given;

mass of tennis ball, m = 60 g = 0.06 kg

height at which the ball fell, h = 2.9 m

percentage of kinetic energy lost, = 23%

The potential energy of the tennis ball = mgh = 0.06 x 9.8 x 2.9

= 1.7052 J

As the ball hits the ground, the potential energy of the ball is converted to kinetic energy.

K.E = 1.7052 J

Kinetic energy lost = 0.23 x 1.7052

= 0.392 J

The remaining kinetic energy ΔK.E = 1.7052 J - 0.392 J

= 1.3132 J

The velocity at which the ball leave the ground is calculated as;

ΔK.E = ¹/₂mv²

`v = \sqrt{(2\Delta K.E)/(m) } \\\\v = \sqrt{(2* 1.3132)/(0.06) }\\\\v = 6.62 \ m/s`

Therefore, the velocity at which the ball leave the ground is 6.62 m/s.