a 0.145 kg pitched baesball moving horizontally at 35 m/s strikes a bat and is popped straight up to a height of 55.6 m before turning around. If the contact time is 0.50 ms, calculate the average force on the ball during contact.

The average force on the ball during the contact is 13,951.9 N.

Explanation:

Given;

mass of the ball, m = 0.145 kg

initial horizontal velocity, uₓ = 35 m/s

the contact time, t = 0.5 ms = 0.5 x 10⁻³ s.

vertical height reached by the ball, h = 55.6 m

The initial vertical velocity of the ball is calculated as;

v² = u² - 2gh

where;

v is the final vertical velocity at maximum height = 0

u is the initial vertical velocity

The resultant velocity is calculated as;

The acceleration of the ball is calculated as;

The average force on the ball during the contact is calculated as;

F = ma

F = (0.145)(96220)

F = 13,951.9 N

Therefore, the average force on the ball during the contact is 13,951.9 N.

e

explanation:

there is no change in the electric potential energy as it is in rest on both conditions.

dang that balls went far