JEE PYQ: Motion in a Plane - Question ID d5dc1e08a1f4 (JEE Main 2022)

In projectile motion on level ground, two key quantities are:

• Range (R): The horizontal distance traveled before the projectile returns to the same vertical level. Formula: $R = \frac{u^2 \sin 2\theta}{g}$
• Maximum height (H): The highest vertical point reached. Formula: $H = \frac{u^2 \sin^2 \theta}{2g}$

The problem states that the range and maximum height are equal, i.e., $R = H$. We are to find $\tan \theta$.

1. Write the expressions for $R$ and $H$: $R = \frac{u^2 \sin 2\theta}{g}, \quad H = \frac{u^2 \sin^2 \theta}{2g}.$ 2. Set $R = H$: $\frac{u^2 \sin 2\theta}{g} = \frac{u^2 \sin^2 \theta}{2g}.$ 3. Cancel the common factor $u^2/g$ (both are positive and nonzero): $\sin 2\theta = \frac{\sin^2 \theta}{2}.$ 4. Recall the double-angle identity $\sin 2\theta = 2\sin\theta\cos\theta$. Substitute: $2\sin\theta\cos\theta = \frac{\sin^2 \theta}{2}.$ 5. Multiply both sides by $2$ to clear the fraction: $4\sin\theta\cos\theta = \sin^2 \theta.$ 6. Divide both sides by $\sin\theta$ (since $\sin\theta \neq 0$ for a nontrivial trajectory): $4\cos\theta = \sin\theta.$ 7. Divide both sides by $\cos\theta$ (assuming $\cos\theta \neq 0$): $4 = \frac{\sin\theta}{\cos\theta} = \tan\theta.$ 8. Therefore, $\tan\theta = 4.$

1. Misremembering formulas: Students sometimes confuse the expressions for range and maximum height. Always write them down explicitly. 2. Skipping the double-angle identity: Forgetting $\sin 2\theta = 2\sin\theta\cos\theta$ can stall the derivation. 3. Canceling $\sin\theta$ prematurely: Ensure $\sin\theta \neq 0$ before dividing; otherwise, you lose a potential solution (though $\theta=0$ is trivial here). 4. Sign errors: Since speeds and angles are positive in this context, no sign issues arise, but in more complex problems, signs matter.

Exam Tip: When range equals maximum height, the relation $\tan\theta = 4$ is a standard result. Memorizing it can save time, but always derive it once to be sure.