JEE PYQ: Motion in a Straight Line - Question ID bc5449bc3fc2 (JEE Main 2022)

Core Formula & Concept:

When a ball is thrown vertically upward or downward, its motion is governed by the equations of uniformly accelerated motion under gravity. The key formulas we use are:

• Displacement as a function of time: $s = ut + \tfrac{1}{2} a t^2$ where $u$ is the initial velocity, $a$ is the acceleration (here, $a = -g$ since we take upward as positive), and $t$ is the time.
• Velocity as a function of time: $v = u + at$
• Maximum height reached above the point of projection: $H = \frac{u^2}{2g}$ (This comes from setting $v = 0$ at the highest point.)

In this problem, the ball is thrown upward from the top of a tower, and we are given the total time until it hits the ground. We must find the height of the tower and then determine the maximum height the ball reaches above the ground.

Step-by-Step Derivation:

Step 1: Define variables and coordinate system

• Let the height of the tower be $h$ (unknown).
• Initial velocity upward: $u = +19.6\ \mathrm{m/s}$.
• Acceleration due to gravity: $a = -g = -9.8\ \mathrm{m/s^2}$ (downward).
• Total time until the ball hits the ground: $t = 6\ \mathrm{s}$.

Step 2: Write the displacement equation

The displacement $s$ of the ball from the top of the tower after time $t$ is: $s = ut + \tfrac{1}{2} a t^2$ Since the ball ends up on the ground, which is $h$ meters below the starting point, we have: $-h = ut + \tfrac{1}{2} a t^2$ Substitute the known values: $-h = (19.6)(6) + \tfrac{1}{2} (-9.8)(6)^2$

Step 3: Calculate the right-hand side

First term: $19.6 \times 6 = 117.6\ \mathrm{m}$
Second term: $\tfrac{1}{2} \times (-9.8) \times 36 = -176.4\ \mathrm{m}$
Sum: $117.6 - 176.4 = -58.8\ \mathrm{m}$
Thus: $-h = -58.8 \implies h = 58.8\ \mathrm{m}$

Step 4: Find the maximum height above the ground

The ball rises to a maximum height above the point of projection (top of the tower) given by: $H_{\text{max above tower}} = \frac{u^2}{2g} = \frac{(19.6)^2}{2 \times 9.8} = \frac{384.16}{19.6} = 19.6\ \mathrm{m}$ Therefore, the maximum height above the ground is: $H_{\text{total}} = h + H_{\text{max above tower}} = 58.8 + 19.6 = 78.4\ \mathrm{m}$

Step 5: Express in the required form and find $k$

The problem states that the maximum height above the ground is $\left(\frac{k}{5}\right)\ \mathrm{m}$. So: $\frac{k}{5} = 78.4 \implies k = 78.4 \times 5 = 392$

Common Traps & Exam Tip:

Students often make the following mistakes:

• Sign confusion: Forgetting to take downward as negative when setting up the displacement equation. Always define a clear coordinate system (upward positive or downward positive) and stick to it.
• Misinterpreting the displacement: The displacement $s$ at time $t$ is the position relative to the starting point. Since the ball ends up on the ground, which is below the starting point, $s$ must be negative if upward is positive.
• Incorrectly calculating the maximum height: Some students add the initial velocity time instead of using $v^2 = u^2 + 2as$ to find the maximum height. Always use the correct kinematic formula for the situation.
• Unit errors: Ensure all units are consistent (meters, seconds, and $\mathrm{m/s^2}$).

Exam Tip: Always draw a diagram showing the initial and final positions, label the known quantities, and write down the kinematic equations before plugging in numbers. This helps avoid sign errors and conceptual misunderstandings.