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JEE PYQ: Motion in a Plane - Question ID 9ad7359ea108 (JEE Main 2019)

ID: 9ad7359ea108JEE Main 2019Single Correct MCQ

Two particles are projected from the same point with the same speed u such that they have the same range R, but different maximum heights, h₁ and h₂. Which of the following is correct ?

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Step-by-step Explanation

Core Formula & Concept:

In projectile motion from level ground, two key results are:

1. Range formula:

R = \frac{u^2 \sin 2\theta}{g}

where

u

is the projection speed,

\theta

the projection angle, and

g

the acceleration due to gravity. 2. Maximum height formula:

h = \frac{u^2 \sin^2 \theta}{2g}

When two projectiles share the same speed $u$ and the same range $R$ , their projection angles must be complementary, i.e. $\theta$ and $90^\circ - \theta$ . This ensures $\sin 2\theta = \sin 2(90^\circ - \theta)$ , so the range is identical.

Step-by-Step Derivation:

Let the two angles be $\theta$ and $90^\circ - \theta$ . Then:

1. Range for both particles:

R = \frac{u^2 \sin 2\theta}{g} = \frac{u^2 \sin 2(90^\circ - \theta)}{g}

2. Maximum heights:

h_1 = \frac{u^2 \sin^2 \theta}{2g}

h_2 = \frac{u^2 \sin^2 (90^\circ - \theta)}{2g} = \frac{u^2 \cos^2 \theta}{2g}

3. Product of heights:

h_1 h_2 = \frac{u^2 \sin^2 \theta}{2g} \cdot \frac{u^2 \cos^2 \theta}{2g} = \frac{u^4 \sin^2 \theta \cos^2 \theta}{4g^2} = \frac{u^4 (\sin \theta \cos \theta)^2}{4g^2}

4. Express $\sin \theta \cos \theta$ in terms of $R$ : Since

\sin 2\theta = 2 \sin \theta \cos \theta

, we have

\sin \theta \cos \theta = \frac{\sin 2\theta}{2} = \frac{Rg}{2u^2}.

5. Substitute back into $h_1 h_2$ :

h_1 h_2 = \frac{u^4}{4g^2} \Bigl(\frac{Rg}{2u^2}\Bigr)^2 = \frac{u^4}{4g^2} \cdot \frac{R^2 g^2}{4u^4} = \frac{R^2}{16}.

6. Rearrange to isolate $R^2$ :

R^2 = 16\,h_1 h_2.

This matches option B.

Common Traps & Exam Tip:

Students often forget that the two angles must be complementary for equal range. They may also confuse the factor of 16 with 4 or 2 by misapplying the double-angle identity. Always write down the complementary-angle relationship explicitly to avoid sign or factor errors.

JEE PYQ: Motion in a Plane - Question ID 9ad7359ea108 (JEE Main 2019)

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Step-by-step Explanation

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