Question

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.

Answer 1

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

Step-by-step 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

`0 = u_y^2 - 2(9.8 * 55.6)\\\\0 = u_y^2 - 1089.76\\\\u_y^2 = 1089.76\\\\u_y = √(1089.76) \\\\u_y = 33.01 \ m/s`

The resultant velocity is calculated as;

`u = √(u_x^2 + u_y^2) \\\\u = √(35^2 + 33.01^2) \\\\u = 48.11 \ m/s`

The acceleration of the ball is calculated as;

`a = (u)/(t) \\\\a = (48.11)/(0.5* 10^(-3)) \\\\a = 96220 \ m/s^2\\\\`

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.