JEE PYQ: Motion in a Plane - Question ID ee74fcab89df (JEE Main 2022)

Core Formula & Concept:

In projectile motion, the maximum height ($H$) attained by a projectile depends on its initial speed ($u$), angle of projection ($\theta$), and acceleration due to gravity ($g$). The formula for maximum height is: $H = \frac{u^2 \sin^2 \theta}{2g}$

Since both projectiles $P_1$ and $P_2$ attain the same height, their maximum heights must be equal. This gives us a relationship between their speeds and angles of projection.

Step-by-Step Derivation:

Let:

• $u_1$ = initial speed of projectile $P_1$
• $u_2$ = initial speed of projectile $P_2$
• $\theta_1$ = angle of projection of $P_1$ (to be found)
• $\theta_2 = 60^\circ$ = angle of projection of $P_2$

Given that the ratio of speeds is: $\frac{u_1}{u_2} = \frac{\sqrt{3}}{\sqrt{2}}$

Since both projectiles reach the same height: $H_1 = H_2 \implies \frac{u_1^2 \sin^2 \theta_1}{2g} = \frac{u_2^2 \sin^2 \theta_2}{2g}$ The $2g$ cancels out: $u_1^2 \sin^2 \theta_1 = u_2^2 \sin^2 \theta_2$

Substitute the speed ratio: $\left(\frac{u_1}{u_2}\right)^2 \sin^2 \theta_1 = \sin^2 \theta_2 \implies \left(\frac{\sqrt{3}}{\sqrt{2}}\right)^2 \sin^2 \theta_1 = \sin^2 60^\circ$ Simplify: $\frac{3}{2} \sin^2 \theta_1 = \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{3}{4}$

Solve for $\sin^2 \theta_1$: $\sin^2 \theta_1 = \frac{3/4}{3/2} = \frac{1}{2} \implies \sin \theta_1 = \frac{1}{\sqrt{2}}$ Thus: $\theta_1 = 45^\circ$

Common Traps & Exam Tip:

Trap 1: Students often confuse the maximum height formula with the range formula. Remember, height depends on $\sin^2 \theta$, while range depends on $\sin 2\theta$.

Trap 2: Misapplying the speed ratio is a common mistake. Ensure that the ratio $\frac{u_1}{u_2} = \frac{\sqrt{3}}{\sqrt{2}}$ is squared correctly in the equation.

Exam Tip: When two projectiles reach the same height, equate their maximum height expressions and simplify. This approach avoids unnecessary complexity and directly leads to the answer.