Arubber ball with a mass of 0.145 kg is dropped from rest. from what height (in m) was the ball dropped, if the magnitude of the ball's momentum is 0.800 kg · m/s just before it lands on the ground?

1.55 m

Explanation:

Momentum: This can be defined as the product of  mass of a body and it velocity. the S.I unit of momentum is kgm/s.

Mathematically,

Momentum can be represented as,

M = mv Equation 1

Where m = mass of the body, v = velocity of the body, M = momentum.

Making v the subject of the equation,

v = M/m Equation 2

Given: M = 0.80 kg.m/s, m = 0.145 kg.

Substituting into equation 2,

v = 0.8/0.145

v = 5.52 m/s.

Using the equation of motion,

v² = u² + 2gs Equation 3.

Where v = final velocity of the rubber ball, u = initial velocity of the rubber ball, s = distance, g = acceleration due to gravity.

Given: v = 5.52 m/s, u = 0 m/s, g = 9.81 m/s².

Substituting into equation 2

5.52² = 0² + 2(9.81)s

30.47 = 19.62s

s = 30.47/19.62

s = 1.55 m.

Thus the ball was dropped from a height of 1.55 m

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