Projectile Motion

**Elite_Guard89** · September 16th 2008, 05:55 PM

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?

**mr fantastic** · September 16th 2008, 07:50 PM

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 $\theta$ .

**su-z-q** · October 1st 2008, 12:54 PM

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?

**Greengoblin** · October 1st 2008, 01:10 PM

$\cos(\theta_{max}) = 1$ (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.

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