Answer to Question #100501 in Mechanics | Relativity for Feker

Question

A projectile launch speed is five times its speed at the maximum height,find the launch angle teta?

Answer 1

Let us assume that the launch angle be $\theta$  and projectile launch speed be $v$

then the horizontal component of velocity = $vcos\theta$

and the vertical component of velocity = $vsin\theta$

then the speed of the object at maximum height = $vcos\theta$

It is given that projectile launch speed is five times its speed at the maximum height

⟹ $v=5vcos\theta$

⟹ $cos\theta=1/5$

⟹$\theta=78.5$ ⁰