After falling from rest at a height of 28.7 m, a 0.502 kg ball rebounds upward, reaching a height of 19.8 m. If the contact between ball and ground lasted 2.40 ms, what average force was exerted on the ball

9080 N

Explanation:

Consider the two motions of the ball.

In the downward motion, initial velocity, u, is 0 (because it falls from rest) and the distance is 28.7 m. Using the equation of motion and using g as 9.8 m/s²,

v² = u² + 2as

v² = 0² + 2 × 9.8 × 28.7 = 562.52

v = 19.7 m/s

For the downward motion, the initial velocity is unknown, the final velocity is 0 and initial velocity is desired. g is negative because the motion is upwars.

0² = v² - 2 × 9.8 × 19.8

v² = 388.08

v = 10.7 m/s

The change in momentum = 0.502(10.7 -(23.7)) = 21.7868 kgm/s

The impulse = change in monetum

Ft = 21.7868 kgm/s

But t = 2.4 ms

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here we know that ball will drop down by total distance 3.2 m

so here we know that

now using kinematics

now in the same time it will cover horizontal distance of 0.75 m

so speed of the ball will be

v=ir = 3x15 =45 v