Question

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 ground exerts a +55 N force on the ball. What is the magnitude of the velocity after it hits the ground?

Answer 1

On E2020 the answer is 9.3 m/s

Answer 2

= 9.25 m/s

Step-by-step 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