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??
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?
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]
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.