A 2.40-kilogram mud ball drops from rest at a height of 18.0 m. If the impact between the ball and the ground lasts 0.50 s, what is the magnitude of the average force exerted by the ball on the ground?

The magnitude of the force exerted by the ball on the ground during the 0.5 s of contact = 90.16 N

Explanation:

Given,

Mass of mud ball = m = 2.40 kg

Height the ball is released from = y = 18 m

Total contact time of ball and the ground = t = 0.5 s

The Newton's second law of motion explains that the magnitude of the change of momentum is equal to the magnitude of a body's impulse.

Change in momentum = Magnitude of Impulse

Change in momentum = (final momentum) - (initial momentum)

Since the ball is dropped from rest, initial momentum = 0 kgm/s

But to calculate its final momentum, we need the ball's final velocity before hitting the ground.

Using the equations of motion,

u = initial velocity of the ball = 0 m/s (ball was dropped from rest)

v = final velocity of the ball = ?

g = acceleration due to gravity = 9.8 m/s²

y = vertical distance covered by the ball = 18 m

v² = u² + 2gy

v² = 0² + (2)(9.8)(18)

v² = 352.8

v = 18.78 m/s

Final momentum of the ball = (m)(v)

= (2.4) × (18.78) = 45.08 kgm/s

Change in momentum = 45.08 - 0 = 45.08 kgm/s

Impulse = Ft

Change in momentum = Magnitude of Impulse

45.08 = F × (0.5)

F = (45.08/0.5) = 90.16 N

Hope this Helps!!!

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