A ball is launched as a projectile with initial speed v at an angle theta above the horizontal. Using conservation of energy, find the maximum height h_max of the ball's flight.(adsbygoogle = window.adsbygoogle || []).push({});

Express your answer in terms of v, g, and theta.

My energy equation is as follows:

0.5m(v^2)cos(theta) +0.5m(v^2)sin(theta) = 0.5m(v^2)cos(theta) + mgh_max

v^2cos(theta) + v^2sin(theta) = 2gh_max + v^2cos(theta)

h_max = (v^2sin(theta))/2g

This isn't right, somehow. I know that the only velocity at max height is in the x-direction, so max height should be a sine function, right? Did I forget to include something in my energy equation?

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# Homework Help: Finding the max height of a ball launched as a projectile using work-energy

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