SATHEE: Motion in a Plane - Result Question 35

Motion in a Plane - Result Question 35

38. For angles of projection of a projectile $(45^{\circ} - θ)$ and $(45^{\circ} + θ)$ , the horizontal ranges described by the projectile are in the ratio of

(a) $1 : 3$

(b) $1 : 2$

(d) $1 : 1$

[2006]

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Answer:

Correct Answer: 38. (d)

Solution:

(d) Horizontal range for projection angle

$(45^{\circ} - θ)$ is, $R_{1} = \frac{u^{2} \sin 2 (45 - θ)}{g}$

Horizontal range projection

angle $(45^{\circ} + θ)$ is, $R_{2} = \frac{u^{2} \sin 2 (45 + θ)}{g}$

According to the condition,

$\Rightarrow \frac{R_{1}}{R_{2}} = \frac{u^{2} \sin 2 (45 - θ)}{u^{2} \sin 2 (45 + θ)} = \frac{\sin (90 - 2 θ)}{\sin (90 + 2 θ)}$

$\Rightarrow \frac{R_{1}}{R_{2}} = \frac{\cos 2 θ}{\cos 2 θ} = \frac{1}{1}$

So, $R_{1} : R_{2} : 1 : 1$

The angle of elevation $(ϕ)$ of the highest point of the projectile and the angle of projection $θ$ are related to each other as $\tan ϕ = \frac{1}{2} \tan θ$

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Motion in a Plane - Result Question 35

38. For angles of projection of a projectile (45∘−θ) and (45∘+θ), the horizontal ranges described by the projectile are in the ratio of

Answer:

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38. For angles of projection of a projectile $(45^{\circ} - θ)$ and $(45^{\circ} + θ)$ , the horizontal ranges described by the projectile are in the ratio of