# Calculate the Momentum

Tags:
1. Apr 6, 2015

### Kikakaru212

The problem statement, all variables and given/known data
A rock of mass 1.10kg and volume 283cm3 is irregular shape. A person throws the rock with a launch angle of 39.0° (above the horizontal) over flat level ground from a height of 0.668m above the ground. Ground level is at an elevation of 138m above sea level. The rock goes 7.40 m down range before it hits the ground. The rock bounces off the ground achieving a maximum height of 12.0cm and traveling an additional 18.0cm down range before coming to rest with a thud. Find the momentum of the rock at the instant it leaves the thrower’s hand.

The attempt at a solution
So what I did was V = √2gx (where x is distance) and I got a velocity of 22.54 m/s. I plugged that into the equation P=m*v to get momentum of 24.8 kgm/s However, i'm not sure I did this right at all.

2. Apr 6, 2015

### Simon Bridge

Welcome to PF.
Hint. Look up "ballistics".
the formula you used is for when the rock is dropped a from rest from a height x, when gravity is the only force. Is that what happened here?

Why is it important that the rock is irregular?
Why is the height above sea level mentioned?
What difference would the angle the rock got thrown make?