# Homework Help: Test tomorrow, can't figure this problem out

1. Feb 1, 2006

### finlejb

An object is thrown vertically upward such that it has a speed of 21 m/s when it reaches two thirds of its maximum height above the launch point. Determine this maximum height.

I can't figure what to do since initial velocity isn't given and I don't see a way to solve for it.

2. Feb 1, 2006

### Staff: Mentor

Reframe the problem this way: An object is thrown upward with an initial speed of 21 m/s and rises to a height of h/3. Solve for h.

3. Feb 1, 2006

### Jameson

Oh I like that suggestion Doc Al. Very clever.

I would use a kinematic equation to help get your answer... this one perhaps?

$$v_{f}^2=v_{i}^2+2ad$$

4. Feb 1, 2006

### finlejb

That equation doesn't really help me though, Jameson. It has two variables in it that I don't know the value of.

And I kind of see where you're going, Doc Al, but why h/3? It seems like 2h/3 would be what I'd want to solve for...

5. Feb 2, 2006

### Hootenanny

Staff Emeritus
What doc al is saying is you can find the maximum heigth because at the point $\frac{2}{3}h$ the initial velocity would be 21ms/2 and at maximum height (which is $\frac{1}{3}h$ above that point) you know the ball would come to rest. This would allow you to use Jameson's equation to calculate the maximum height.

6. Feb 2, 2006

### Staff: Mentor

Hootenanny already explained it, but let me put it this way. You have three points of interest:
(1) start: y = 0; v = ?
(2) "mid" point: y = 2h/3; v = 21 m/s
(3) top: y = h; v = 0 m/s

What I'm suggesting is that between points 2 & 3 the object travels a distance of h/3. A perfect opportunity to apply Jameson's kinematic equation.