Momentum - velocity due to height

[ngorecki]

Homework Statement

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

Homework Equations

p = mv

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

 

[haruspex]

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

 

[ngorecki]

Total velocity is conserved...?

 

[haruspex]

How many laws of conservation have you heard of?

 

[ngorecki]

energy and momentum

 

[haruspex]

energy and momentum

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

 

[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

 

[haruspex]

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

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.
Your answer is clearly wrong since it exceeds the initial velocity.