# object thrown vertically

by kimikims
Tags: object, thrown, vertically
 P: 37 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??
Mentor
P: 40,241
 Quote by kimikims V^2-Vo^2 = 2g(1/3H)
Right (where Vo = 0).

 I know I need to plug in the numbers, but how do I get all of the equation to one side, to find H??
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?
 HW Helper P: 2,274 You got $$V^2 = V_{o}^2 + 2g\frac{2}{3}H$$ where V is 58 m/s and $$0 = V_{o}^2 + 2gH$$
HW Helper
P: 2,274

## object thrown vertically

Doc Al, how come initial speed = 0 ?
Mentor
P: 40,241
 Quote by Cyclovenom Doc Al, how come initial speed = 0 ?
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.
HW Helper
P: 2,274
 Quote by Doc Al 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.
Oh Yes, i wasn't thinking about it

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