JEE PYQ: Motion in a Plane - Question ID a715e5e1133e (JEE Main 2012)

When a particle is projected with an initial velocity $u$ at an angle $\theta$ with the horizontal, its motion can be resolved into two independent components:

• Vertical motion: The maximum height $H$ reached by the projectile is given by $H = \frac{u^2 \sin^2 \theta}{2g}$ where $g$ is the acceleration due to gravity.
• Horizontal motion: The horizontal range $R$ is given by $R = \frac{u^2 \sin 2\theta}{g}$

The key insight is that the same initial speed $u$ governs both the maximum height and the maximum range. The maximum range occurs when $\sin 2\theta = 1$, i.e., $\theta = 45^\circ$, yielding $R_{\max} = \frac{u^2}{g}$

1. From the given maximum height $H = 10$ m, use the vertical-motion formula: $10 = \frac{u^2 \sin^2 \theta_{\text{vert}}}{2g}$ For maximum height, the stone is thrown straight up, so $\theta_{\text{vert}} = 90^\circ$ and $\sin 90^\circ = 1$. Thus $10 = \frac{u^2}{2g} \quad \Longrightarrow \quad u^2 = 20g.$

2. The maximum horizontal range occurs at $\theta = 45^\circ$. Substitute $u^2 = 20g$ into the range formula: $R_{\max} = \frac{u^2 \sin 90^\circ}{g} = \frac{20g \cdot 1}{g} = 20\text{ m}.$

Many students mistakenly assume that the maximum height and maximum range are numerically equal or related by a factor of $\sqrt{2}$. They may also confuse the angle for maximum height ($90^\circ$) with the angle for maximum range ($45^\circ$). Always remember:

• Maximum height uses $\theta = 90^\circ$.
• Maximum range uses $\theta = 45^\circ$.
• The two formulas share the same $u^2$, so once you find $u^2$ from the height, you can directly compute the range.