JEE PYQ: Motion in a Plane - Question ID 93fd5cd93bfb (JEE Main 2025)

Core Formula & Concept:

In projectile motion, the trajectory of a particle projected at an angle $\theta$ with the horizontal can be analyzed by resolving its initial velocity into horizontal ($v_{0x}$) and vertical ($v_{0y}$) components:

• $v_{0x} = v_0 \cos \theta$ (constant, as no acceleration in horizontal direction)
• $v_{0y} = v_0 \sin \theta$ (changes due to gravity, $g = 10 \, \text{m/s}^2$ downward)

The vertical motion is governed by the equations of uniformly accelerated motion:

• Velocity at time $t$: $v_y(t) = v_{0y} - gt$
• Displacement at time $t$: $y(t) = v_{0y} t - \frac{1}{2} g t^2$

The time to reach maximum height ($t_{\text{max}}$) is when $v_y = 0$:

$t_{\text{max}} = \frac{v_{0y}}{g} = \frac{v_0 \sin \theta}{g}$

The height traversed in a given time interval is the difference in vertical displacement between the start and end of that interval.

Step-by-Step Derivation:

Given:

• Initial speed, $v_0 = 60 \, \text{m/s}$
• Projection angle, $\theta = 30^\circ$
• Acceleration due to gravity, $g = 10 \, \text{m/s}^2$

Step 1: Compute initial vertical velocity

$v_{0y} = v_0 \sin \theta = 60 \cdot \sin 30^\circ = 60 \cdot 0.5 = 30 \, \text{m/s}$

Step 2: Compute time to reach maximum height

$t_{\text{max}} = \frac{v_{0y}}{g} = \frac{30}{10} = 3 \, \text{seconds}$

Step 3: Compute height traversed in the first second ($h_0$)

Height traversed in the first second is the vertical displacement from $t = 0$ to $t = 1$:

$h_0 = y(1) - y(0) = \left(v_{0y} \cdot 1 - \frac{1}{2} g \cdot 1^2\right) - 0$ $h_0 = 30 \cdot 1 - \frac{1}{2} \cdot 10 \cdot 1 = 30 - 5 = 25 \, \text{m}$

Step 4: Compute height traversed in the last second before reaching maximum height ($h_1$)

The "last second" before reaching maximum height is from $t = 2$ to $t = 3$ (since $t_{\text{max}} = 3$ s).

Compute vertical displacement at $t = 2$ and $t = 3$:

$y(2) = 30 \cdot 2 - \frac{1}{2} \cdot 10 \cdot 2^2 = 60 - 20 = 40 \, \text{m}$ $y(3) = 30 \cdot 3 - \frac{1}{2} \cdot 10 \cdot 3^2 = 90 - 45 = 45 \, \text{m}$

Height traversed in the last second:

$h_1 = y(3) - y(2) = 45 - 40 = 5 \, \text{m}$

Step 5: Compute the ratio $h_0 : h_1$

$h_0 : h_1 = 25 : 5 = 5 : 1$ Common Traps & Exam Tip:

Trap 1: Misidentifying the "last second". Students often assume the last second is from $t = 2.5$ to $t = 3.5$, which is incorrect because the particle reaches maximum height at $t = 3$ s. The last second is from $t = 2$ to $t = 3$.

Trap 2: Forgetting that height traversed is the difference in vertical displacement, not the displacement itself. Some students mistakenly use $y(1)$ or $y(2)$ directly as $h_0$ or $h_1$.

Exam Tip: Always compute $t_{\text{max}}$ first. Then, for any time interval, compute $y(t)$ at the start and end of the interval and take the difference. This avoids confusion about "height traversed".