Question

A ball is projected vertically up with a velocity of $40 \text{ ms}^{-1}$ from ground. At the same time another ball is dropped from a height of $100 \text{ m}$. The magnitudes of their velocities are equal after

• A. $1 \text{ s}$
• B. $2 \text{ s}$
• C. $3 \text{ s}$
• D. $4 \text{ s}$

Hint:

In many competitive exams, if $g$ is not specified, using $10 \text{ ms}^{-2}$ instead of $9.8 \text{ ms}^{-2}$ often leads to cleaner integer results that match the options provided.

Answer 1

The Correct Option is B

We need to find the time '$t$' at which the absolute values of the velocities of both balls are identical. Let $g = 10 \text{ ms}^{-2}$ for simplicity. 1. Velocity of the Projected Ball ($v_1$): Using $v = u - gt$ (upward motion): $$v_1 = 40 - 10t$$

2. Velocity of the Dropped Ball ($v_2$): Using $v = u + gt$ where $u = 0$ (downward motion): $$v_2 = 0 + 10t = 10t\lt strong\gt 3. Equating Magnitudes:\lt /strong\gt |v_1| = |v_2|$$ $$40 - 10t = 10t$$ (Note: Since we are looking for a point where both are still in motion, $v_1$ is still positive/upward). $$40 = 20t \implies t = \frac{40}{20} = 2 \text{ s}$$ At $t = 2 \text{ s}$, the first ball has a velocity of $40 - 20 = 20 \text{ ms}^{-1}$ (upward), and the second ball has a velocity of $20 \text{ ms}^{-1}$ (downward). Their magnitudes are equal.