JEE PYQ: Motion in a Plane - Question ID fd69b6b6b15a (JEE Main 2023)

Core Formula & Concept:

In projectile motion, the trajectory of a ball thrown near the Earth's surface can be analyzed by resolving its motion into horizontal ($x$) and vertical ($y$) components. The key formulas and concepts involved are:

• Maximum Vertical Height ($H$): When a ball is thrown vertically upward, its initial velocity is purely vertical. The maximum height $H$ is reached when the final vertical velocity becomes zero. Using the kinematic equation: $v_y^2 = u_y^2 - 2gH$ At maximum height, $v_y = 0$, so: $0 = u_y^2 - 2gH \implies H = \frac{u_y^2}{2g}$
• Maximum Horizontal Range ($R$): When the ball is thrown at an angle $\theta$ to the horizontal, the horizontal range is given by: $R = \frac{u^2 \sin(2\theta)}{g}$ The maximum range occurs when $\sin(2\theta) = 1$, i.e., $\theta = 45^\circ$. Thus: $R_{\text{max}} = \frac{u^2}{g}$
• Relationship Between $H$ and $R_{\text{max}}$: From the above, we can relate the maximum height and maximum range. For the same initial speed $u$: $H = \frac{u^2}{2g} \quad \text{and} \quad R_{\text{max}} = \frac{u^2}{g}$ Thus: $R_{\text{max}} = 2H$

This relationship is crucial because it connects the maximum vertical height to the maximum horizontal range without needing to know the initial velocity or the acceleration due to gravity explicitly.

--- Step-by-Step Derivation:

Given:

• Maximum vertical height, $H = 136$ m.

We need to find the maximum horizontal distance ($R_{\text{max}}$) the man can throw the same ball.

1. Determine the initial velocity for vertical throw: Using the formula for maximum height: $H = \frac{u_y^2}{2g}$ Here, $u_y$ is the initial vertical velocity. Solving for $u_y$: $u_y^2 = 2gH \implies u_y = \sqrt{2gH}$
2. Relate initial velocity to maximum range: The maximum horizontal range occurs when the ball is thrown at $45^\circ$ to the horizontal. The initial velocity $u$ is the same in both cases (vertical throw and $45^\circ$ throw). Thus: $u = u_y = \sqrt{2gH}$ The maximum range is: $R_{\text{max}} = \frac{u^2}{g}$ Substituting $u^2 = 2gH$: $R_{\text{max}} = \frac{2gH}{g} = 2H$
3. Calculate the maximum horizontal distance: Substituting $H = 136$ m: $R_{\text{max}} = 2 \times 136 = 272 \text{ m}$

Thus, the maximum horizontal distance the man can throw the ball is 272 m, which corresponds to option B.

--- Common Traps & Exam Tip:

Students often make the following mistakes in this question:

• Assuming the same initial velocity for vertical and horizontal throws: Some students incorrectly assume that the initial velocity for the vertical throw is the same as the horizontal component of velocity in a projectile motion. However, the initial velocity $u$ is the same in both cases, but its components differ based on the angle of projection.
• Ignoring the angle for maximum range: The maximum range occurs at $45^\circ$, not at $90^\circ$ (vertical throw) or $0^\circ$ (horizontal throw). Students sometimes forget this and incorrectly calculate the range.
• Misapplying the range formula: The range formula $R = \frac{u^2 \sin(2\theta)}{g}$ is often misapplied by using $\sin(\theta)$ instead of $\sin(2\theta)$. This leads to incorrect results.
• Forgetting the relationship between $H$ and $R_{\text{max}}$: The key insight is that $R_{\text{max}} = 2H$. Students who do not recall this relationship may attempt to solve the problem using unnecessary steps, increasing the chance of errors.

Exam Tip: Always remember that for a given initial speed, the maximum horizontal range is twice the maximum vertical height. This relationship simplifies the problem significantly and avoids lengthy calculations.