A 2.00-kilogram mud ball drops from rest at a height of 17.0 m. If the impact between the ball and the ground lasts 0.46 s, what is the magnitude of the average force exerted by the ball on the ground

79.35 N

Explanation: according to the impulse momentum theorem,

Impulse = change in momentum

Where impulse = force × time and change in momentum = m ( v - u).

The object was initially at rest, hence it initial velocity is zero.

To get the final velocity, we use the formulae below

v² = u² + 2gh

Where h = height of the cliff = 17m

v² = 2 × 9.8 × 17

v² = 333.2

v = √333.2

v = 18.25 m/s

At t = 0.46s and v = 18.25 m/s, we can get the average force of impact

F×0.46 = 2 (18.25 - 0)

F × 0.46 = 2 (18.25)

F × 0.46 = 36.5

F = 36.5 /0.46 = 79.35 N

representative si the final answer! : )