Question

Two balls P and Q are thrown vertically upwards simultaneously from the ground with velocities $20\ \text{m s}^{-1}$ and $35\ \text{m s}^{-1}$ respectively. The distance between the two balls when the velocity of P becomes zero is: (g = $10\ \text{m s}^{-2}$)}

• A. 30 m
• B. 20 m
• C. 50 m
• D. 70 m

Hint:

Always compute positions at the same time instant for relative distance problems.

Answer 1

The Correct Option is A

Step 1: For ball P, time to reach maximum height is when velocity becomes zero: \[ v = u - gt = 0 \Rightarrow t = \frac{20}{10} = 2\ \text{s} \]
Step 2: Height of P at $t = 2$ s: \[ s_P = ut - \frac{1}{2}gt^2 = 20(2) - \frac{1}{2}(10)(2^2) \] \[ = 40 - 20 = 20\ \text{m} \]
Step 3: Height of Q at same time: \[ s_Q = 35(2) - \frac{1}{2}(10)(4) \] \[ = 70 - 20 = 50\ \text{m} \]
Step 4: Distance between balls: \[ \Delta s = s_Q - s_P = 50 - 20 = 30\ \text{m} \] Hence, separation is 30 m.