# Homework Help: Momentum - velocity due to height

1. Nov 14, 2012

### ngorecki

1. The problem statement, all variables and given/known data

A 0.20 kg ball is thrown straight up into the air with an initial speed of 19 m/s. Find the momentum of the ball at the following locations.

(a) at its maximum height
(b) halfway to its maximum height

Variable:
ball = .2kg
Vi = 19 m/s

2. Relevant equations

p = mv

3. The attempt at a solution

a) since the object is at max height the velocity is 0. therefore momentum is also 0.
b)This is where I ger lost...
I tried using the equation for momentum but I am not sure on how to solve for velocity.
My teacher said something about height being = to v^2 but he said this quickly and not sure what the usefulness of that is...

It would be greatly appreciated if I could get help ASAP

2. Nov 14, 2012

### haruspex

During ascent, gravity acts on the ball. As a result, the ball's vertical momentum is not conserved. What is conserved?

3. Nov 14, 2012

### ngorecki

Total velocity is conserved...?

4. Nov 14, 2012

### haruspex

How many laws of conservation have you heard of?

5. Nov 14, 2012

### ngorecki

energy and momentum

6. Nov 14, 2012

### haruspex

Good. I've explained why you can't using conservation of momentum here, so that leaves ....?

7. Nov 15, 2012

### ngorecki

Alright so
mgh = 1/2mv^2
Cancel mass
gh = 1/2v^2
9.8(h) = .5(19)^2
h = 1768.9

Then mgh(.5h) = 1/2mv^2
Cancel mass
9.8(.5*1768.9) = 1/2v^2
8667.61 = 1/2v^2
17335.22 = v^2
131.66 = v

8. Nov 15, 2012

### haruspex

I strongly urge you to get into the habit of working with the algebra as long as possible, only plugging numbers in right at the end. It will help you avoid mistakes, help spot mistakes, and help others follow what you're doing.