How Do You Calculate the Maximum Height of an Object Thrown Vertically Upward?

[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??

 

[Doc Al]

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]

You got

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

where V is 58 m/s

and

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

 

[Pyrrhus]

Doc Al, how come initial speed = 0 ?

 

[Doc Al]

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]

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