Two bodies are projected at angles θ and (90 ∘ −θ) to the horizontal with the same speed. The ratio of their times of flight is

1. Break Down the Problem

We have two bodies projected at angles $\theta$ and $(90^\circ - \theta)$ with the same speed $u$. We need to find the ratio of their times of flight.

2. Relevant Concepts

The time of flight $T$ for a projectile launched at an angle $\theta$ can be given by the formula: $T = \frac{2u \sin \theta}{g}$ where $g$ is the acceleration due to gravity.

3. Analysis and Detail

For the first body projected at an angle $\theta$: $T_1 = \frac{2u \sin \theta}{g}$

For the second body projected at an angle $(90^\circ - \theta)$: $T_2 = \frac{2u \sin (90^\circ - \theta)}{g}$ Using the trigonometric identity $\sin (90^\circ - \theta) = \cos \theta$, we have: $T_2 = \frac{2u \cos \theta}{g}$

4. Verify and Summarize

Now we can find the ratio of the times of flight $\frac{T_1}{T_2}$: $\text{Ratio} = \frac{T_1}{T_2} = \frac{\frac{2u \sin \theta}{g}}{\frac{2u \cos \theta}{g}} = \frac{\sin \theta}{\cos \theta} = \tan \theta$

Final Answer

The ratio of the times of flight is given by: $\frac{T_1}{T_2} = \tan \theta$