Maximum Height of Projectile Using Cons. of Energy

[akbro93704]

Homework Statement

A projectile is launched with a speed of 45 m/s at an angle of 56° above the horizontal. Use conservation of energy to find the maximum height reached by the projectile during its flight.

Homework Equations

PE=KE
mgh=1/2mv^2
gh=1/2v^2
h=.5v^2/g

The Attempt at a Solution

I thought in this equation that the angle wouldn't matter, but I know I'm wrong. I thought you could just plug it the numbers to the manipulated formula h=.5v^2/g. Please help! thanks

 

[hage567]

Yes the angle matters. You need to consider the vertical component of the initial velocity.

 

[baba89]

Your approach is on the right track, but you are missing a key factor - the initial vertical velocity of the projectile. In order to use the equation h=.5v^2/g, you need to use the vertical component of the initial velocity, which is v0sinθ, where θ is the launch angle. So the correct equation would be h=.5(v0sinθ)^2/g. This will give you the maximum height reached by the projectile.