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Yuri [45]
4 years ago
12

A 0.330-kg volleyball is thrown vertically downward with a speed of 0.150 m/s in a place where g = 9.81 m/s2. It takes it 0.0655

s to reach the ground. What is the magnitude of its momentum just before it hits the ground?
Physics
1 answer:
Law Incorporation [45]4 years ago
8 0

Answer:

The magnitude of its momentum just before it hits the ground is 0.2607\frac{kg\,m}{s}

Explanation:

Magnitude of momentum is:

p=mv (1)

with p the momentum, v the velocity and m the mass.

For an object with constant acceleration we can use the kinematic equation (2) to determine its velocity at any time

v=vo+at (2)

with v the final velocity, vo the initial velocity, a the acceleration and t the time. We can use the values vo=0.150\frac{m}{s} , a=g=9.81\frac{m}{s^{2}} t=0.0655 s on (2) to find the velocity before it reaches the ground

v=(0.150)+(9.81)(0.0655) = 0.79 \frac{m}{s}

With that velocity we can use equation (1) to find the momentum of the ball just before it hits the ground:

p=(0.330)(0.79)

p=0.2607\frac{kg\,m}{s}

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