Physics: After falling from rest at a height of 32.3 m, a 0.556 kg ball rebounds upward, reaching a height of 22.1 m. if the contact between ball and ground lasted 1.62 ms, what average force was exerted on the ball?

After falling from rest at a height of 32.3 m,

tina7659

12.03.2022

F = 15771.6 N

Explanation:

Initial velocity of ball just before it will collide is given as

$v_i = \sqrt{2gh_1}$

$v_i = \sqrt{2(9.81)(32.2)}$

v_i = 25.13 m/s

now for final speed of rebound we have

$v_f = \sqrt{2gh_2}$

$v_f = \sqrt{2(9.81)(22.1)}$

v_f = 20.82 m/s

now the average force is given as

$F = \frac{mv_f - mv_i}{\Delta t}$

$F = \frac{0.556(20.82 + 25.13)}{1.62 \times 10^{-3}}$

F = 15771.6 N

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honeyaranda326pay2bl

03.05.2023

The two pink or Rose colored ones are parallelograms

Step-by-step explanation:

They both have lines that are one pair of parallel lines to each other

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