Motion in a Plane - Result Question 35
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38. For angles of projection of a projectile $(45^{\circ}-\theta)$ and $(45^{\circ}+\theta)$, the horizontal ranges described by the projectile are in the ratio of
======= ####38. For angles of projection of a projectile $(45^{\circ}-\theta)$ and $(45^{\circ}+\theta)$, the horizontal ranges described by the projectile are in the ratio of
3e0f7ab6f6a50373c3f2dbda6ca2533482a77bed:content/english/neet-pyq-chapterwise/physics/motion-in-a-plane/motion-in-a-plane—result-question-35.md (a) $1: 3$
(b) $1: 2$
(c) $2: 1$
(d) $1: 1$
[2006]
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Answer:
Correct Answer: 38. (d)
Solution:
- (d) Horizontal range for projection angle
$(45^{\circ}-\theta)$ is, $R_1=\frac{u^{2} \sin 2(45-\theta)}{g}$
Horizontal range projection
angle $(45^{\circ}+\theta)$ is, $R_2=\frac{u^{2} \sin 2(45+\theta)}{g}$
According to the condition,
$\Rightarrow \frac{R_1}{R_2}=\frac{u^{2} \sin 2(45-\theta)}{u^{2} \sin 2(45+\theta)}=\frac{\sin (90-2 \theta)}{\sin (90+2 \theta)}$
$\Rightarrow \frac{R_1}{R_2}=\frac{\cos 2 \theta}{\cos 2 \theta}=\frac{1}{1}$
So, $R_1: R_2: 1: 1$
The angle of elevation $(\phi)$ of the highest point of the projectile and the angle of projection $\theta$ are related to each other as $\tan \phi=\frac{1}{2} \tan \theta$
