Question

Two bodies are projected at angles $\theta $ and $\left(\left(90\right)^{o} - \theta \right)$ to the horizontal with the same speed. The ratio of their times of flight is

• A. $sin \theta : 1$
• B. $cos \theta : 1$
• C. $sin \theta :cos ⁡ \theta $
• D. $cos \, \theta :sin ⁡ \theta $

Answer 1

Answer: (C)

Time of flight, $T=\frac{2 u \, sin \theta }{g}$
For $\theta , \, T_{1}=\frac{2 u \, sin \theta }{g}$ ......(i)
For $(90^{o} - \theta ), \, T_{2}=\frac{2 u \, sin \left(\left(90\right)^{o} - \theta \right)}{g}=\frac{2 u \, cos ⁡ \theta }{g}$ .....(ii)
Dividing Equation. (i) by Equation. (ii), we get
$\therefore \, \, \, \frac{T_{1}}{T_{2}}=\frac{2 u \, sin \theta }{g}\times \frac{g}{2 u \, cos ⁡ \theta }=\frac{sin ⁡ \theta }{cos ⁡ \theta }$