Posted by **jenny** on Sunday, September 17, 2006 at 4:28pm.

A stone is thrown vertically upward. On its way up it passes point A with speed, v and point B, 3 m higher than A, with speed 1/2v.

Calculate a) the speed v and b) the max height reached by the stone above point B.

I can't find the equation

You need to derive the solution in problems such as this. You can do that by "thinking it through", using equations you should already know. There are many ways to solve the problem. I will choose energy.

In going from point A to point B, the potential energy increases by M g H, there h = 3 m and g = 9.8 m/s^2.

The change (increase) in potential anergy will equal the decrease in kinetic energy. Therefore

M g H = (1/2) M [v^2 - (v/2)^2]

2 g H = (3/4) v^2

Solve that for v. You can use conservation of energy again to solve for the maximum height attained. That would occur when the potential energy increase equals the initial kinetic energy.

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