JEE PYQ: Motion in a Plane - Question ID df2358655f14 (JEE Main 2022)

Core Formula & Concept:

In projectile motion, the range ($R$) and maximum height ($H$) of a projectile launched from ground level are determined by its initial velocity ($v_0$) and launch angle ($\theta$). The key formulas are:

• Range of a projectile: $R = \frac{v_0^2 \sin(2\theta)}{g}$ where $g$ is the acceleration due to gravity.
• Maximum height of a projectile: $H = \frac{v_0^2 \sin^2(\theta)}{2g}$

The maximum range occurs when $\sin(2\theta) = 1$, i.e., at $\theta = 45^\circ$. At this angle, the range formula simplifies to: $R_{\text{max}} = \frac{v_0^2}{g}$

The question states that the person can throw the ball to a maximum range of 100 m. This implies that the launch angle is $45^\circ$, and the initial velocity $v_0$ is such that: $R_{\text{max}} = 100 = \frac{v_0^2}{g}$

We are asked to find the maximum height the same ball can reach when thrown vertically upward (i.e., at $\theta = 90^\circ$). This is a special case of projectile motion where the entire initial velocity is directed upward.

Step-by-Step Derivation:

Step 1: Relate maximum range to initial velocity.

Given the maximum range: $R_{\text{max}} = 100 = \frac{v_0^2}{g}$ We solve for $v_0^2$: $v_0^2 = 100g$

Step 2: Find maximum height for vertical throw.

When the ball is thrown vertically upward, the maximum height $H$ is given by: $H = \frac{v_0^2}{2g}$ Substitute $v_0^2 = 100g$ from Step 1: $H = \frac{100g}{2g} = 50 \text{ m}$

Step 3: Verify using projectile height formula.

Alternatively, using the general height formula for $\theta = 90^\circ$: $H = \frac{v_0^2 \sin^2(90^\circ)}{2g} = \frac{v_0^2 (1)}{2g} = \frac{v_0^2}{2g}$ Again, substituting $v_0^2 = 100g$: $H = \frac{100g}{2g} = 50 \text{ m}$

Conclusion:

The maximum height the person can throw the ball is 50 m, which corresponds to option B.

Common Traps & Exam Tip:

• Misinterpreting "maximum range": Some students assume the given range is for an arbitrary angle, not the maximum range. The key is recognizing that 100 m is the maximum possible range, which occurs only at $45^\circ$. This directly gives $v_0^2 = R_{\text{max}} \cdot g$.
• Confusing range and height formulas: Students often mix up the formulas for range ($R = \frac{v_0^2 \sin(2\theta)}{g}$) and height ($H = \frac{v_0^2 \sin^2(\theta)}{2g}$). Remember that range depends on $\sin(2\theta)$, while height depends on $\sin^2(\theta)$.
• Ignoring the vertical throw case: The question asks for the height when the ball is thrown vertically. Some students mistakenly use the height formula for $\theta = 45^\circ$, leading to incorrect results. Always read the question carefully to identify the correct launch angle.
• Algebraic errors: When substituting $v_0^2 = 100g$ into the height formula, students sometimes forget to cancel $g$ or misplace the factor of 2. Double-check each step to avoid such mistakes.

Exam Tip: For problems involving projectile motion, always start by writing down the known formulas and identifying which variables are given. Here, recognizing that the given range is the maximum range is crucial to solving the problem efficiently.