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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the answer would be c

the working equation for the angular size is the product of the physical size and 360 degrees divided by the product of two pi and distance from earth. the minimum size would be 0.559 degrees while the maximum size would be 0.49 degrees.