Q102 / 200 · NEET 2006 · Physics
For angles of projection of projectile at angle $\left(45^{\circ}-\theta\right)$ and $\left(45^{\circ}+\theta\right)$, the horizontal range described by the projectile are in the ratio of:
- A$2: 1$
- Correct answerB$1: 1$
- C2:3
- D$1: 2$
Solution
Horizontal range
$$
=R=\frac{u^2 \sin ^2 \theta}{g}
$$
When angle of projection is $\left(45^{\circ}-\theta\right)$ then
$$
\begin{aligned}
R_1 & =\frac{u^2 \sin 2\left(45^{\circ}-\theta\right)}{g} \\
& =\frac{u^2 \sin \left(90^{\circ}-\theta\right)}{g} \\
& =\frac{u^2 \cos 2 \theta}{g}
\end{aligned}
$$
Now $\quad \frac{R_1}{R_2}=\frac{u^2 \cos \frac{2 \theta}{8}}{u^2 \cos \frac{2 \theta}{g}}=1: 1$
Hence, the correct answer is option: 'B'.