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?
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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 $\displaystyle = U \cos \theta$ (since the vertical component of the velocity is equal to zero at maximum height).Quote:
Originally Posted by Elite_Guard89
Therefore $\displaystyle U = 5U \cos \theta$. Solve for $\displaystyle \theta$.
i am now confused wat?(Doh)Quote:
Originally Posted by mr fantastic
$\displaystyle \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, $\displaystyle v\cos(\theta_{max}) = v$, an, unknown initial velocity, $\displaystyle u=5v\cos(\theta)$, and an unknown angle $\displaystyle \theta$, which you need to find. I'd say there isn't enough information to find it. if you are assuming gravity of $\displaystyle -9.8ms^{-2}$, 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.
