JEE PYQ: Motion in a Plane - Question ID 2f6934af2fb1 (JEE Main 2025)

Core Formula & Concept:
The problem deals with projectile motion, specifically focusing on the maximum height achieved by a particle. In projectile motion, when a particle is launched with an initial velocity $u$ at an angle $\theta$ with respect to the *horizontal* axis, the motion can be resolved into horizontal and vertical components. The vertical motion is governed by gravity. The initial vertical component of the velocity is $u_y = u \sin \theta$. At the maximum height ($h_m$), the vertical component of the velocity momentarily becomes zero. Using the kinematic equation $v_y^2 = u_y^2 + 2a_y s_y$ for the vertical motion, where $v_y = 0$, $a_y = -g$ (taking upward direction as positive), and $s_y = h_m$: $0^2 = (u \sin \theta)^2 + 2(-g)h_m$ $0 = u^2 \sin^2 \theta - 2gh_m$ Rearranging this equation, we get the standard formula for maximum height: $h_m = \frac{u^2 \sin^2 \theta}{2g}$ Here, $u$ is the initial speed of projection, $\theta$ is the angle of projection with the *horizontal*, and $g$ is the acceleration due to gravity.
Step-by-Step Derivation:
1. Understand the given angle: The problem states that the angle of projection of the particle is measured from the *vertical axis* as $\phi$. This is crucial. In most standard projectile motion formulas, the angle is measured from the horizontal. 2. Relate $\phi$ to the standard horizontal angle $\theta$: Let the initial speed of projection be $u$. If $\phi$ is the angle with the vertical axis, then the standard angle of projection with the horizontal axis, $\theta$, can be found from the geometric relationship: $\theta + \phi = 90^\circ$ Therefore, the angle with the horizontal is: $\theta = 90^\circ - \phi$ 3. Substitute $\theta$ into the maximum height formula: Now, we substitute this expression for $\theta$ into the standard maximum height formula derived above: $h_m = \frac{u^2 \sin^2 \theta}{2g}$ Substitute $\theta = (90^\circ - \phi)$: $h_m = \frac{u^2 \sin^2 (90^\circ - \phi)}{2g}$ 4. Apply trigonometric identity: We use the trigonometric identity $\sin (90^\circ - X) = \cos X$. Applying this identity for $X = \phi$: $\sin (90^\circ - \phi) = \cos \phi$ Therefore, $\sin^2 (90^\circ - \phi) = (\cos \phi)^2 = \cos^2 \phi$. 5. Final expression for $h_m$: Substitute this back into the equation for $h_m$: $h_m = \frac{u^2 \cos^2 \phi}{2g}$ This expression gives the maximum height $h_m$ as a function of the initial speed $u$, the angle $\phi$ (measured from the vertical), and the acceleration due to gravity $g$. Comparing this derived expression with the given options, the correct option corresponds to: $h_m = \frac{u^2 \cos^2 \phi}{2g}$
Common Traps & Exam Tip:
1. Incorrect Angle Reference: The most frequent mistake students make is directly using $\phi$ as the angle with the horizontal, leading to $h_m = \frac{u^2 \sin^2 \phi}{2g}$. Always pay close attention to whether the angle is given with respect to the horizontal or the vertical axis. A quick sketch can clarify this relationship. 2. Trigonometric Identity Error: Some might incorrectly use $\cos(90^\circ - \phi) = \sin \phi$ or make other trigonometric errors. Remember $\sin(90^\circ - X) = \cos X$ and $\cos(90^\circ - X) = \sin X$. 3. Missing Square: Forgetting to square the sine or cosine term, leading to $h_m = \frac{u^2 \cos \phi}{2g}$, is another common error. Exam Tip: Whenever angles are specified in a projectile motion problem, it's always good practice to draw a small diagram showing the initial velocity vector, the horizontal axis, and the vertical axis. Mark the given angle and then deduce the angle with the horizontal (if not directly given) before applying any standard formulas. This avoids misinterpretation and helps in correctly applying trigonometric identities.