Question

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

Answer 1

`cos^-1(1/5) ~~ 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#