JEE PYQ: Motion in a Straight Line - Question ID d81de6ff9d33 (JEE Main 2014)

In this problem, we analyze the motion of a particle under constant gravitational acceleration ($g$) when thrown vertically upward from a tower of height $H$. The key concepts and formulas involved are:

• Equations of Motion: For uniformly accelerated motion along a straight line, the displacement ($s$), initial velocity ($u$), acceleration ($a$), and time ($t$) are related by: $s = ut + \frac{1}{2} a t^2$ Here, $a = -g$ (since upward is taken as positive and gravity acts downward).
• Time to Reach Highest Point ($t_1$): At the highest point, the final velocity becomes zero. Using $v = u + at$: $0 = u - g t_1 \implies t_1 = \frac{u}{g}$
• Total Time to Hit the Ground ($t_2$): The particle is thrown upward, reaches the highest point, then falls back down past the tower to the ground. The total displacement from the point of projection to the ground is $-H$ (negative because it ends below the starting point). Using the displacement equation: $-H = u t_2 - \frac{1}{2} g t_2^2$ This is a quadratic in $t_2$.
• Relation Between Times: The problem states that the total time to hit the ground ($t_2$) is $n$ times the time to reach the highest point ($t_1$): $t_2 = n t_1 = n \frac{u}{g}$

Step 1: Express $t_1$ (time to reach highest point)

At the highest point, velocity $v = 0$. Using $v = u - g t_1$: $0 = u - g t_1 \implies t_1 = \frac{u}{g}$

Step 2: Express $t_2$ (total time to hit the ground)

The particle is projected upward from height $H$ and lands on the ground, so the net displacement is $-H$. Using the displacement equation: $-H = u t_2 - \frac{1}{2} g t_2^2$ Rearrange: $\frac{1}{2} g t_2^2 - u t_2 - H = 0$ This is a quadratic equation in $t_2$: $g t_2^2 - 2 u t_2 - 2 H = 0$

Step 3: Use the given relation $t_2 = n t_1$

Substitute $t_1 = \frac{u}{g}$ into $t_2 = n t_1$: $t_2 = n \frac{u}{g}$

Step 4: Substitute $t_2$ into the quadratic equation

Plug $t_2 = \frac{n u}{g}$ into $g t_2^2 - 2 u t_2 - 2 H = 0$: $g \left(\frac{n u}{g}\right)^2 - 2 u \left(\frac{n u}{g}\right) - 2 H = 0$ Simplify: $g \cdot \frac{n^2 u^2}{g^2} - \frac{2 n u^2}{g} - 2 H = 0$ $\frac{n^2 u^2}{g} - \frac{2 n u^2}{g} - 2 H = 0$ Multiply through by $g$ to eliminate denominators: $n^2 u^2 - 2 n u^2 - 2 g H = 0$ Factor out $u^2$: $u^2 (n^2 - 2 n) - 2 g H = 0$ Rearrange: $2 g H = u^2 (n^2 - 2 n)$ Factor the right side: $2 g H = n u^2 (n - 2)$

Step 5: Match with the given options

The derived relation is: $2 g H = n u^2 (n - 2)$ This matches Option C.

• Sign Convention: Many students confuse the sign of displacement and acceleration. Always take upward as positive and downward (including gravity) as negative. The displacement to the ground is $-H$, not $H$.
• Quadratic Equation Setup: Students often incorrectly set up the displacement equation for $t_2$ as $H = u t_2 - \frac{1}{2} g t_2^2$, forgetting that the particle ends below the starting point. This leads to the wrong relation.
• Time Relation Misinterpretation: The phrase "time taken to hit the ground is $n$ times that taken to reach the highest point" is sometimes misread as $t_1 = n t_2$, reversing the relation. Always verify which time is larger.
• Algebraic Simplification: In the final steps, students may fail to factor $n^2 - 2n$ as $n(n - 2)$, missing the match with Option C. Practice factoring quadratics to avoid this.

Exam Tip: When solving such problems, always:

1. Define a clear sign convention.
2. Write down known values and relations explicitly.
3. Substitute carefully, especially when dealing with quadratic equations.
4. Cross-check units and dimensions in the final relation to eliminate absurd options.