Question

Two bodies are projected at angles $\theta$ and $({90}^{\circ }-\theta )$ 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

The correct option is C $\sin \theta :\cos \theta$

$\text{Step 1: Use expression for time of flight}$

Let $u$ be the speed of projection.

$\text{For first body:}$

Angle of projection is $\theta$.

So, Time of flight $T_1=\frac{2u\sin \theta }{g}$ $....\left(1\right)$

$\text{For second body:}$

Angle of projection is ${90}^o-\theta$.

where $u$ is the speed of projection.

Time of flight $T_2=\frac{2u\sin ({90}^o-\theta )}{g}=\frac{2u\cos \theta }{g}$ $....\left(2\right)$

$Step2:Calculateratioof$ $\frac{T_1}{T_2}$

Using equations $\left(1\right)$ and $\left(2\right)$, The ratio of time of flight $\frac{T_1}{T_2}=\frac{\sin \theta }{\cos \theta }$

Hence, Option C correct