# Projectile Motion

• Sep 16th 2008, 05:55 PM
Elite_Guard89
Projectile Motion
I have been stuck on this for a while....

A projectile's launch speed is five times it speed at maximum height. Find the launch angle.

Anybody have any clue how to do this?
• Sep 16th 2008, 07:50 PM
mr fantastic
Quote:

Originally Posted by Elite_Guard89
I have been stuck on this for a while....

A projectile's launch speed is five times it speed at maximum height. Find the launch angle.

Anybody have any clue how to do this?

Speed at maximum height $= U \cos \theta$ (since the vertical component of the velocity is equal to zero at maximum height).

Therefore $U = 5U \cos \theta$. Solve for $\theta$.
• Oct 1st 2008, 12:54 PM
su-z-q
Quote:

Originally Posted by mr fantastic
Speed at maximum height $= U \cos \theta$ (since the vertical component of the velocity is equal to zero at maximum height).

Therefore $U = 5U \cos \theta$. Solve for $\theta$.

i am now confused wat?(Doh)
• Oct 1st 2008, 01:10 PM
Greengoblin
$\cos(\theta_{max}) = 1$ (at maximum height), since this is the parabolic path of a projectile. So you have an unknwn velocity at maximum height, $v\cos(\theta_{max}) = v$, an, unknown initial velocity, $u=5v\cos(\theta)$, and an unknown angle $\theta$, which you need to find. I'd say there isn't enough information to find it. if you are assuming gravity of $-9.8ms^{-2}$, and have the final velocity, the launch and impact heights, then you could plug in the values for x and y, and solve simultaneous equations using the equations of motion.