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Object thrown vertically

by kimikims
Tags: object, thrown, vertically
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kimikims
#1
Sep24-04, 02:07 PM
P: 36
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??
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Doc Al
#2
Sep24-04, 02:24 PM
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Quote 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?
Pyrrhus
#3
Sep24-04, 02:26 PM
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You got

[tex] V^2 = V_{o}^2 + 2g\frac{2}{3}H [/tex]

where V is 58 m/s

and

[tex] 0 = V_{o}^2 + 2gH [/tex]

Pyrrhus
#4
Sep24-04, 02:28 PM
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Object thrown vertically

Doc Al, how come initial speed = 0 ?
Doc Al
#5
Sep24-04, 02:35 PM
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Quote 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.
Pyrrhus
#6
Sep24-04, 02:36 PM
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Quote 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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