Question

It was calculated that a shell when fired from a gun with a certain velocity and at an angle of elevation $\frac{5 \pi }{3 6}$ rad should strike a given target. In actual practice, it was found that a hill just prevented the trajectory. At what angle of elevation should the gun be fired to hit the target ?

• A. $\frac{5 \pi }{3 6}\text{rad}$
• B. $\frac{1 1 \pi }{3 6}\text{rad}$
• C. $\frac{7 \pi }{3 6}\text{rad}$
• D. $\frac{1 3 \pi }{3 6}\text{rad}$

Answer 1

Answer: (D)

Ranges for complementary angles are the same
$R_{\theta } = R_{\left(\right. \frac{\pi }{2} - \theta \left.\right)}$
$\therefore $ Required angle $=\frac{\pi }{2}-\frac{5 \pi }{3 6}=\frac{1 3 \pi }{3 6}rad$