Q154 / 225 · JEE Main 2004 · Physics
A projectile can have the same range $R$ for two angles of projection. If $T_1$ and $T_2$ be the time of flights in the two cases, then the product of the two time of flights is directly proportional to
- A$1 / R^2$
- B$1 / \mathrm{R}$
- Correct answerC$\mathrm{R}$
- D$\mathrm{R}^2$
Solution
Range is same for complimentary angles. $\mathrm{T}_1=\frac{2 \mathrm{u} \sin \theta}{\mathrm{g}}$ and $\mathrm{T}_2=\frac{2 \mathrm{u} \sin (90-\theta)}{\mathrm{g}}$ and $R=\frac{u^2 \sin 2 \theta}{\mathrm{g}}$
$$
\therefore \mathrm{T}_1 \mathrm{~T}_2=\frac{2 \mathrm{u} \sin \theta}{\mathrm{g}} \times \frac{2 \mathrm{u} \cos \theta}{\mathrm{g}}=\frac{2 \mathrm{R}}{\mathrm{g}} \text {. }
$$
Hence, the correct answer is option: 'C'.