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

1 Answer
Mar 11, 2018

cos1(15)78

Explanation:

If the angle of projection is α and the projection speed is u, the projectile starts off with the horizontal and vertical velocity components ucosα and usinα, 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 ucosα.

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