Question

a baseball with mass of 730.0 grams drops from a 4 meter height and hits the ground ad bounces high to 2.5 meter height. if total contact time of ball with the ground is 5.0m/s, then find average exerted force from the ground on the ball during contact time?

Answer 1

Initial velocity when the ball is dropped, u = 0
Distance travelled, s = 4 m
Acceleration, a = 9.81 m/s²
Using the formula, v² - u² = 2as,
Where v is the velocity with which the ball hit the ground.
v = SQRT[2as]
= SQRT[2 x 9.81 x 4]
= 8.86 m/s

After hitting the ground, the ball bounces to a height h = 2.5 m
Consider the initial velocity as u'
Final velocity, v' = 0
Acceleration, a = - 9.81 m/s²
Using the formula, v'² - u'² = 2ah,
0 - u'² = 2 x (- 9.81) x 2.5
u' = SQRT[2 x 9.81 x 2.5]
= 7.00 m/s

Change in momentum of the ball, P = mV
= m [v - (-u')]                [u' is negative compared with the direction of v]
= m [v + u']
= 0.73 x [8.86 + 7.00]
= 11.58 kg m/s

Average force = P/t
= 11.58 / (5 x 10^-3)
= 2.32 x 10³ N