JEE PYQ: Motion in a Plane - Question ID 77ad1024026b (JEE Main 2021)

In projectile motion, two key quantities are the range ($R$) and the maximum height ($H$).

• Range formula: $R = \frac{u^2 \sin 2\theta}{g}$ where $u$ is the initial speed, $\theta$ is the projection angle, and $g$ is the acceleration due to gravity.
• Maximum height formula: $H = \frac{u^2 \sin^2 \theta}{2g}$

A crucial observation is that the range depends on $\sin 2\theta$, which is symmetric about $\theta = 45^\circ$. Specifically, for any angle $\theta$ and its complement $90^\circ - \theta$, the ranges are equal because $\sin 2\theta = \sin 2(90^\circ - \theta)$. Here, $42^\circ$ and $48^\circ$ are complementary angles ($42^\circ + 48^\circ = 90^\circ$), so their ranges must be identical.

Step 1: Establish the range equality

Given $\theta_1 = 42^\circ$ and $\theta_2 = 48^\circ$, compute $\sin 2\theta_1 = \sin 84^\circ,\quad \sin 2\theta_2 = \sin 96^\circ.$ But $\sin 96^\circ = \sin (180^\circ - 84^\circ) = \sin 84^\circ$. Therefore $R_1 = \frac{u^2 \sin 84^\circ}{g} = R_2 = \frac{u^2 \sin 96^\circ}{g}.$ Hence $R_1 = R_2$.

Step 2: Compare the maximum heights

Compute $H_1 = \frac{u^2 \sin^2 42^\circ}{2g},\quad H_2 = \frac{u^2 \sin^2 48^\circ}{2g}.$ Since $\sin 48^\circ > \sin 42^\circ$ (because $48^\circ$ is closer to $90^\circ$), it follows that $H_2 > H_1.$

Step 3: Match with the given options

We have shown $R_1 = R_2$ and $H_1 < H_2$. This corresponds exactly to option B.

Many students mistakenly believe that the larger angle always gives the larger range. They overlook the symmetry of the sine function around $45^\circ$. Always remember:

• For complementary angles ($\theta$ and $90^\circ - \theta$), the ranges are equal.
• The maximum height increases as the angle moves from $0^\circ$ toward $90^\circ$.

Double-check the angle values and use the symmetry property to save time in the exam.