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muminat
3 years ago
14

A 0.060 kg ball hits the ground with a speed of –32 m/s. The ball is in contact with the ground for 45 milliseconds and the grou

nd exerts a +55 N force on the ball. What is the magnitude of the velocity after it hits the ground?
Physics
2 answers:
FinnZ [79.3K]3 years ago
7 0

<u>Answer</u>

= 9.25 m/s


<u>Explanation</u>

The Newton's second law of motion states that, the change in momentum is directly propotional to the force producing it and it takes place in the direction of force.

F = ma

f = m(v-u)/t

ft = m(v-u)

∴ 55 × 45/1000 = 0.060(v - -32)

2.475 = 0.06(v + 32)

2.475/0.6 = v + 32

41.25 = v + 32

v = 41.25 -32

= 9.25 m/s

Tanzania [10]3 years ago
4 0

On E2020 the answer is 9.3 m/s

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