Object thrown vertically

1. Sep 24, 2004

kimikims

An object is thrown vertically upward such that it has a speed of 58 m/s when it reaches two thirds of it's maximum height above the launch point.
The acceleration of gravity is 9.8 m/s^2

Find the maximum height h. Answer in units of m.

I know:

V: 58 m/s at y=2/3 H
H = ?
V = 0

V^2-Vo^2 = 2g(1/3H)

I know I need to plug in the numbers, but how do I get all of the equation to one side, to find H??

2. Sep 24, 2004

Staff: Mentor

Right (where Vo = 0).

Divide or multiply both sides to eliminate what's in front of H.

For example, if you had an equation: 7x^2 = 67bZ, and you wanted to isolate Z, just divide both sides by 67b: (7x^2)/67b = (67bZ)/67b = Z; so Z = (7x^2)/67b. Make sense?

3. Sep 24, 2004

Pyrrhus

You got

$$V^2 = V_{o}^2 + 2g\frac{2}{3}H$$

where V is 58 m/s

and

$$0 = V_{o}^2 + 2gH$$

Last edited: Sep 24, 2004
4. Sep 24, 2004

Pyrrhus

Doc Al, how come initial speed = 0 ?

5. Sep 24, 2004

Staff: Mentor

If you look at the motion on the way down, then the object starts from rest and falls a distance of H/3 to reach a speed of 58 m/s.

6. Sep 24, 2004

Pyrrhus

Oh Yes, i wasn't thinking about it