JEE PYQ: Motion in a Straight Line - Question ID 6dbf0a9a7b1a (JEE Main 2020)

When an object moves with constant acceleration, its motion can be described using the kinematic equations. The key formulas we will use are:

• Displacement as a function of time: $s = ut + \frac{1}{2} a t^2$
• Velocity as a function of time: $v = u + at$

Here, $u$ is the initial velocity, $a$ is the acceleration, $t$ is the time, and $s$ is the displacement.

In this problem:

• The helicopter starts from rest and accelerates upward with acceleration $g$. This means its initial velocity is $0$, and its acceleration is $+g$ (taking upward as positive).
• When the helicopter reaches height $h$, a food packet is dropped. At that instant, the packet has the same velocity as the helicopter (since it was inside the helicopter).
• Once dropped, the packet is under the influence of gravity alone, so it experiences acceleration $-g$ (downward).

We need to find the time taken by the packet to reach the ground after being dropped from height $h$.

Step 1: Find the velocity of the helicopter (and packet) at height $h$

The helicopter starts from rest ($u = 0$) and accelerates upward with acceleration $g$. Let $t_1$ be the time taken by the helicopter to reach height $h$. Using the displacement formula:

Solving for $t_1$:

Now, the velocity $v$ of the helicopter (and packet) at height $h$ is:

So, when the packet is dropped, it has an initial upward velocity $v = \sqrt{2gh}$.

Step 2: Motion of the packet after being dropped

After being dropped, the packet moves under gravity with:

• Initial velocity: $u = \sqrt{2gh}$ (upward)
• Acceleration: $a = -g$ (downward)
• Displacement: $s = -h$ (since it moves from height $h$ to the ground, which is downward)

Using the displacement formula:

Substituting the values:

Simplify:

Multiply both sides by $-1$:

Rearrange:

Multiply through by $2$ to simplify:

Divide the entire equation by $g$ to make it dimensionless:

Let $\tau = \sqrt{\frac{h}{g}}$. Then $\frac{h}{g} = \tau^2$, and the equation becomes:

This is a quadratic equation in $t$:

Solve for $t$ using the quadratic formula $t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a = 1$, $b = -2 \sqrt{2} \tau$, and $c = -2 \tau^2$:

Simplify the discriminant:

Thus:

We discard the negative root because time cannot be negative:

Substitute back $\tau = \sqrt{\frac{h}{g}}$:

Calculate the numerical value of $2 + \sqrt{2}$:

Thus:

This matches option A.

Students often make the following mistakes in this problem:

1. Ignoring the initial velocity of the packet: Many assume the packet is dropped from rest, leading to the incorrect formula $t = \sqrt{\frac{2h}{g}}$. However, the packet inherits the velocity of the helicopter at height $h$, which is $\sqrt{2gh}$.
2. Sign errors in acceleration: Some students take upward as negative or downward as positive, leading to incorrect signs in the kinematic equations. Always define a consistent coordinate system (e.g., upward as positive).
3. Solving the quadratic equation incorrectly: The quadratic equation has two roots, but only the positive root is physically meaningful. Students sometimes pick the wrong root or make algebraic errors in solving the quadratic.
4. Approximation errors: The exact solution is $t = (2 + \sqrt{2}) \sqrt{\frac{h}{g}}$, and $2 + \sqrt{2} \approx 3.414$. Students may miscalculate this or confuse it with other numerical approximations.

Exam Tip: Always break the problem into two parts:

1. Motion of the helicopter until the packet is dropped.
2. Motion of the packet after being dropped.