# A projectile's launch speed is five times its speed at maximum height.Find launch angle.How do I find this?

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Mar 11, 2018

#### Answer:

${\cos}^{-} 1 \left(\frac{1}{5}\right) \approx {78}^{\circ}$

#### Explanation:

If the angle of projection is $\alpha$ and the projection speed is $u$, the projectile starts off with the horizontal and vertical velocity components $u \cos \alpha$ and $u \sin \alpha$, respectively. Of these, the horizontal component of the velocity is constant throughout the motion, while the vertical component keeps on changing at a constant rate $- g$.

At the topmost point of the path, the projectile's vertical velocity component vanishes - and thus its velocity at that point is exactly horizontal, and its magnitude is $u \cos \alpha$.

Thus \$u = 5u cos alpha implies cos alpha = 1/5#

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