JEE PYQ: Motion in a Plane - Question ID 9883a2c51350 (JEE Main 2026)

Core Formula & Concept: This problem deals with projectile motion, which is a classic application of kinematics under constant acceleration. When a particle is projected into the air, its motion can be analyzed independently in horizontal and vertical directions.

1. Horizontal Motion: Assuming negligible air resistance, there is no acceleration in the horizontal direction. Thus, the horizontal component of velocity ($v_x$) remains constant throughout the motion. $v_x = u_x$ $x = u_x t$
2. Vertical Motion: In the vertical direction, the particle experiences a constant downward acceleration due to gravity, $g$. We typically choose an upward direction as positive and downward as negative. The equations of motion for constant acceleration apply: $v_y = u_y - gt$ $y = u_y t - \frac{1}{2}gt^2$ $v_y^2 = u_y^2 - 2gy$
3. Components of Initial Velocity: If a particle is projected with an initial speed $u$ at an angle $\theta$ from the horizontal, its initial velocity components are: $u_x = u \cos\theta$ $u_y = u \sin\theta$
4. Time to Attain Maximum Height: At the maximum height of its trajectory, the vertical component of the velocity ($v_y$) becomes momentarily zero. Using the first equation of motion for vertical direction: $0 = u_y - gt_{max} \implies t_{max} = \frac{u_y}{g}$ Step-by-Step Derivation: We are given the initial projection angle, the time to reach maximum height, and the time to land on the building. We need to find the height $H$ of the building.
Step 1: Determine the initial vertical component of velocity ($u_y$). The problem states that the ball attains maximum height in $t_{max} = 2$ s. At maximum height, the vertical component of its velocity, $v_y$, is zero. We can use the first equation of motion for vertical displacement: $v_y = u_y - g t_{max}$ Given $g = 10 \text{ m/s}^2$ and $t_{max} = 2$ s: $0 = u_y - (10 \text{ m/s}^2)(2 \text{ s})$ $u_y = 20 \text{ m/s}$
Step 2: Utilize the projection angle (optional but good for complete understanding). The ball is thrown at $45^{\circ}$ from the horizontal. We know $u_y = u \sin 45^{\circ}$. So, $20 = u \left(\frac{1}{\sqrt{2}}\right) \implies u = 20\sqrt{2} \text{ m/s}$. The initial horizontal component of velocity $u_x = u \cos 45^{\circ} = (20\sqrt{2})\left(\frac{1}{\sqrt{2}}\right) = 20 \text{ m/s}$. For this problem, $u_x$ is not directly required to find $H$, but it's consistent with $u_y$ since $\sin 45^{\circ} = \cos 45^{\circ}$.
Step 3: Calculate the height $H$ of the building. The ball lands on the building in $T = 3$ s after launch. The height $H$ is the vertical displacement $y$ at this time $T$. We use the second equation of motion for vertical displacement: $y = u_y T - \frac{1}{2} g T^2$ Substitute the values we have: $u_y = 20 \text{ m/s}$, $T = 3$ s, and $g = 10 \text{ m/s}^2$. $H = (20 \text{ m/s})(3 \text{ s}) - \frac{1}{2} (10 \text{ m/s}^2)(3 \text{ s})^2$ $H = 60 \text{ m} - \frac{1}{2} (10 \text{ m/s}^2)(9 \text{ s}^2)$ $H = 60 \text{ m} - (5 \text{ m/s}^2)(9 \text{ s}^2)$ $H = 60 \text{ m} - 45 \text{ m}$ $H = 15 \text{ m}$
The height of the building $H$ is 15 m. The final answer is $\boxed{\text{15}}$. Common Traps & Exam Tip:
1. Misinterpreting Time of Flight: A common mistake is to confuse the time to land on the building (3 s) with the total time of flight to return to the initial launch height (which would be $2 \times t_{max} = 4$ s if it landed at the same level). Since the ball lands on a roof of height $H$, the motion is not symmetric, and the total time of flight to the ground could be even longer. Always use the given time for the specific event described.
2. Incorrect Sign Conventions: Ensure consistent sign conventions for vertical motion. If upward is taken as positive, then initial vertical velocity $u_y$ is positive, and acceleration due to gravity $g$ must be taken as negative ($-g$). If downward is positive, $u_y$ would be negative, and $g$ would be positive. I've used $y = u_y t - \frac{1}{2}gt^2$, where $g$ is taken as the magnitude of acceleration, and the minus sign accounts for its downward direction when $u_y$ is upward positive.
3. Assuming Landing at Max Height: Some students might erroneously assume that the ball lands on the building at its maximum height. The problem states it attains maximum height in 2 s and lands in 3 s, clearly indicating these are different points in time and thus different positions.
4. Unnecessary Calculations: While determining the initial speed $u$ and horizontal velocity $u_x$ can be done, they are not strictly necessary to find $H$. Focus on isolating the relevant information for the specific unknown variable. In this case, $u_y$ and time $T$ are sufficient for $H$.