Answer:
The velocity at which the ball leave the ground is 6.62 m/s.
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
[tex]v = \sqrt{\frac{2\Delta K.E}{m} } \\\\v = \sqrt{\frac{2\times 1.3132}{0.06} }\\\\v = 6.62 \ m/s[/tex]
Therefore, the velocity at which the ball leave the ground is 6.62 m/s.
