JEE PYQ: Motion in a Straight Line - Question ID d3665b48dd14 (JEE Main 2004)

When a ball is released from rest under the influence of gravity, it undergoes uniformly accelerated motion in a straight line. The key equations governing this motion are derived from Newton’s second law and kinematic relations. The fundamental formulas we use are:

• Displacement as a function of time: $s(t) = ut + \frac{1}{2} a t^2$ where $s(t)$ is the displacement at time $t$, $u$ is the initial velocity, and $a$ is the acceleration.
• Velocity as a function of time: $v(t) = u + at$
• Relation between displacement, velocity, and acceleration (without time): $v^2 = u^2 + 2as$

In this problem, the ball is released from rest, so $u = 0$. The acceleration is due to gravity, $a = g$, acting downward. We take downward as the positive direction for simplicity. The total height of the tower is $h$, and the total time taken to reach the ground is $T$.

Our goal is to find the position of the ball at time $t = \frac{T}{3}$, i.e., how far it has fallen in one-third of the total time.

Step 1: Express total displacement in terms of $T$ and $g$

Since the ball starts from rest and falls a distance $h$ in time $T$, we use the displacement formula: $h = ut + \frac{1}{2} g T^2$ But $u = 0$, so: $h = \frac{1}{2} g T^2 \quad \Rightarrow \quad g = \frac{2h}{T^2}$ This gives us a relation between $g$, $h$, and $T$.

Step 2: Find displacement at $t = \frac{T}{3}$

Let $s$ be the distance fallen in time $t = \frac{T}{3}$. Again, using the displacement formula: $s = ut + \frac{1}{2} g t^2 = 0 + \frac{1}{2} g \left(\frac{T}{3}\right)^2 = \frac{1}{2} g \cdot \frac{T^2}{9} = \frac{g T^2}{18}$ Now substitute $g = \frac{2h}{T^2}$ from Step 1: $s = \frac{1}{18} \cdot \frac{2h}{T^2} \cdot T^2 = \frac{2h}{18} = \frac{h}{9}$ So, the ball has fallen $\frac{h}{9}$ meters in $\frac{T}{3}$ seconds.

Step 3: Determine position from the ground

The tower has total height $h$. If the ball has fallen $s = \frac{h}{9}$ meters, then its position from the ground is: $\text{Position} = h - s = h - \frac{h}{9} = \frac{8h}{9}$

Step 4: Match with given options

The position from the ground at $t = \frac{T}{3}$ is $\frac{8h}{9}$ meters. This matches option A.

Trap 1: Misinterpreting "position from the ground"
Many students calculate the distance fallen ($s$) and directly choose an option like $\frac{h}{9}$, which is the distance fallen, not the position from the ground. Always remember: position from the ground = total height − distance fallen.

Trap 2: Incorrect sign convention
Some students take upward as positive and downward as negative, leading to confusion in displacement. Consistency is key. Here, taking downward as positive simplifies the math.

Trap 3: Assuming constant velocity
A common misconception is that the ball moves with constant speed. In reality, it accelerates due to gravity, so distance is not proportional to time. The $t^2$ dependence is crucial.

Exam Tip:
Always relate total time and total distance first to find $g$ in terms of $h$ and $T$. This avoids needing the numerical value of $g$ and keeps the solution general and clean.