A0.060 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 ground exerts a +55 n force on the ball. what is the magnitude of the velocity after it hits the ground?

Answer

= 9.25 m/s

Explanation

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

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