JEE PYQ: Motion in a Plane - Question ID 5cf4d1820c2b (JEE Main 2024)

Core Formula & Concept:

In projectile motion, the maximum height ($H$) reached by a projectile launched vertically or at an angle depends on its initial vertical velocity component. The key formula for maximum height when air resistance is neglected is:

$H = \frac{u_y^2}{2g}$ where:

• $H$ is the maximum height,
• $u_y$ is the initial vertical component of velocity,
• $g$ is the acceleration due to gravity (approximately $9.8 \mathrm{~m/s^2}$, but often taken as $10 \mathrm{~m/s^2}$ in JEE for simplification).

When the projectile is launched at an angle $\theta$ with the horizontal, the vertical component of the initial velocity is: $u_y = u \sin \theta$ where $u$ is the initial speed (magnitude of velocity).

Thus, the maximum height can also be expressed as: $H = \frac{(u \sin \theta)^2}{2g} = \frac{u^2 \sin^2 \theta}{2g}$

This shows that the maximum height is proportional to the square of the initial speed, assuming the launch angle remains unchanged.

Step-by-Step Derivation:

Step 1: Express the original maximum height

Given that the original maximum height is $64 \mathrm{~m}$, we write: $H_1 = \frac{u^2 \sin^2 \theta}{2g} = 64 \quad \text{(1)}$

Step 2: Halve the initial velocity

The new initial velocity is: $u_2 = \frac{u}{2}$

Step 3: Compute the new maximum height

Using the same formula, the new maximum height $H_2$ is: $H_2 = \frac{u_2^2 \sin^2 \theta}{2g} = \frac{\left(\frac{u}{2}\right)^2 \sin^2 \theta}{2g} = \frac{u^2 \sin^2 \theta}{4 \cdot 2g} = \frac{u^2 \sin^2 \theta}{8g}$

Step 4: Relate $H_2$ to $H_1$

From equation (1), we know: $\frac{u^2 \sin^2 \theta}{2g} = 64 \Rightarrow \frac{u^2 \sin^2 \theta}{8g} = \frac{64}{4} = 16$

Step 5: Conclude the new maximum height

Therefore: $H_2 = 16 \mathrm{~m}$ Common Traps & Exam Tip:

Trap 1: Assuming height is directly proportional to velocity. Many students mistakenly think that if velocity is halved, height is also halved. However, since height depends on the square of velocity, halving the velocity reduces the height to one-fourth, not half.

Trap 2: Ignoring the angle. Even if the projectile is launched at an angle, the maximum height depends only on the vertical component. Since the angle is unchanged when velocity is scaled, the $\sin^2 \theta$ factor cancels out in the ratio. So, the result holds regardless of the launch angle.

Exam Tip: Always remember that in projectile motion, range is proportional to $u^2$, and maximum height is also proportional to $u^2$. So, any change in initial speed affects these quantities quadratically.

Thus, when initial velocity is halved, maximum height becomes one-fourth of the original. $64 \mathrm{~m} \div 4 = 16 \mathrm{~m}$, which matches the correct answer.