Question

A projectile's launch speed is 2 times its speed at maximum height. Find launch angle ${\theta }_0$.

• A. 30°
• B. 37°
• C. 45°
• D. 60°

Answer 1

The correct option is D

60°

Let the projectile be thrown with speed U in the direction ${\theta }_0$.

At highest point the vertical component of its velocity becomes 0, so it can't go any higher. But the horizontal component still remains constant as V cos $\theta$ as there is no acceleration in the x-direction to either slow down or increase the velocity. Now according to question
Launch speed V is twice speed at maximum height
$\Rightarrow$ v = 2v cos $\theta$
$\Rightarrow$cos $\theta$ = $\frac{1}{2}$
$\Rightarrow$$\theta$ = ${60}^{\circ }$.