Answer:
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
[tex]0 = u_y^2 - 2(9.8 \times 55.6)\\\\0 = u_y^2 - 1089.76\\\\u_y^2 = 1089.76\\\\u_y = \sqrt{1089.76} \\\\u_y = 33.01 \ m/s[/tex]
The resultant velocity is calculated as;
[tex]u = \sqrt{u_x^2 + u_y^2} \\\\u = \sqrt{35^2 + 33.01^2} \\\\u = 48.11 \ m/s[/tex]
The acceleration of the ball is calculated as;
[tex]a = \frac{u}{t} \\\\a = \frac{48.11}{0.5\times 10^{-3}} \\\\a = 96220 \ m/s^2\\\\[/tex]
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.
