A ball is launched with initial speed v from ground level up a frictionless slope. The slope makes an angle theta with the horizontal. Using conservation of energy, find the maximum vertical height hmax to which the ball will climb. Express your answer in terms of v, g, and theta. You may or may not use all of these variables in your answer.(adsbygoogle = window.adsbygoogle || []).push({});

The max height of a body is given by

mgh = 0.5mv^2

gh = 0.5v^2

h = v^2/2g

since the ball is at an angle, and at max height there is zero velocity in the y direction, the only velocity is that in the x-direction, or vcos(theta).

v^2cos(theta)/2g = hmax

but that isn't right. Anyone have any ideas?

3. The attempt at a solution

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# Maximum height reached by a ball using work-energy theorem

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