JEE PYQ: Motion in a Straight Line - Question ID c4c0fbb8f9c4 (JEE Main 2005)

This problem involves motion in a straight line under two distinct phases:

1. Free-fall (no friction): The parachutist falls freely under gravity for the first 50 m. During this phase, acceleration is constant at $g = 9.8\ m/s^2$ downward. We use the kinematic relation $v^2 = u^2 + 2\,a\,s$ to relate initial speed $u$, final speed $v$, acceleration $a$, and displacement $s$. Here $u = 0$, $a = g$, and $s = 50$ m.
2. Decelerated motion (with parachute): After the parachute opens, the parachutist decelerates at $2\ m/s^2$ until reaching the ground with speed $3\ m/s$. We again use $v^2 = u^2 + 2\,a\,s$ but now $a = -2\ m/s^2$ (deceleration), $v = 3\ m/s$, and $u$ is the speed just after the parachute opens.

By combining the two phases, we can find the total height from which the parachutist bailed out.

Phase 1: Free fall of 50 m

Initial speed $u_1 = 0$, acceleration $a_1 = g = 9.8\ m/s^2$, displacement $s_1 = 50$ m.
Using $v^2 = u^2 + 2\,a\,s$, the speed at the end of the free-fall phase is $v_1^2 = 0 + 2 \times 9.8 \times 50 = 980 \quad\Longrightarrow\quad v_1 = \sqrt{980} \approx 31.30\ m/s.$

Phase 2: Decelerated descent

The parachutist now decelerates at $a_2 = -2\ m/s^2$, starting from $u_2 = v_1 \approx 31.30\ m/s$, and reaches the ground with $v_2 = 3\ m/s$.
Using $v_2^2 = u_2^2 + 2\,a_2\,s_2$, we solve for the displacement $s_2$ during this phase: $$3^2 = (31.30)^2 + 2 \times (-2)\times s_2 \quad\Longrightarrow\quad 9 = 980 - 4\,s_2 \quad\Longrightarrow\quad 4\,s_2 = 980 - 9 = 971 \quad\Longrightarrow\quad s_2 = \frac{971}{4} = 242.75\ m.$$

Total height

The total height $H$ from which the parachutist bailed out is the sum of the two displacements: $H = s_1 + s_2 = 50 + 242.75 = 292.75\ m.$ Rounding to the nearest whole number gives $H \approx 293$ m.

1. Sign of acceleration: Many students forget that deceleration means negative acceleration in the chosen coordinate system. If you take $a = +2\ m/s^2$ instead of $-2\ m/s^2$, you will get a nonsensical negative displacement. 2. Exact vs approximate $g$: Using $g = 10\ m/s^2$ instead of $9.8\ m/s^2$ gives $v_1 = \sqrt{1000} \approx 31.62$, leading to $s_2 = 245$ and total height $295$ m, which is close but not among the options. Stick to $g = 9.8$ for precision. 3. Unit consistency: Ensure all units are in meters and seconds; mixing cm or km will ruin the calculation.