Question

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?

Answer 1

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:

$p=(0.330)(0.79)$

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