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

1. Mar 19, 2007

### ph123

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.

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?

2. Mar 19, 2007

### PhanthomJay

Looks like you forgot to square the trig functions. At the top of the flight, for example, the KE is
$$KE = 1/2m(vcos\theta)^2$$. Also, in the initial case, you don't have to break v into its x and y components. You can just use 1/2mv^2 and get the same result.

3. Mar 19, 2007

### ph123

thanks! i got

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

You actually cleared up a big confusion for me in general. I always was using v^2trig(theta) in lots of problems and never understood what was going wrong.

4. Mar 19, 2007

### PhanthomJay

Yes, that is correct, and better than my suggestion to not break up v_initial into its components. In which case you would get, not to confuse you,
$$1/2mv^2 = mgh_{max} + 1/2m(vcos\theta)^2$$
$$v^2 = 2gh_{max} + (vcos\theta)^2$$
$$h_{max} = (v^2 - (vcos\theta)^2 )/2g$$
$$h_{max} = (v^2(1 - (cos\theta)^2)/2g$$
$$h_{max} = (v^2sin^2\theta)/2g$$
which is a bit more tedious than your approach.