Question

The total speed of a projectile at its greatest height is $\sqrt{\frac{6}{7}}$ of its total speed when it is at half of its greatest height. What is the angle of projection?

• A. $\theta ={30}^o$
• B. $\theta ={45}^o$
• C. $\theta ={60}^o$
• D. $\theta ={75}^o$

Answer 1

The correct option is A $\theta ={30}^o$

The total velocity at the highest point is $v_o\cos \theta$
At half the greatest height $\left(\frac{h}{2}\right)$ the resultant of $\frac{(v_o\sin \theta {)}^2}{2}$ and $v_o\cos \theta$ is the velocity.
$\therefore \sqrt{\frac{6}{7}}\sqrt{\frac{v_o^2+v_o^2{\cos }^2\theta }{2}}=v_o\cos \theta$
Solving ${\cos }^2\theta =\frac{3}{4}$ cos $\theta =\left(\frac{\sqrt{3}}{2}\right)$
$\theta ={30}^{\circ }$