Q 11-03-005NEETNEET 2021Top questionMedium
A car starts from rest and accelerates at $5\ \text{m/s}^2$. At $t = 4$ s, a ball is dropped out of a window by a person sitting in the car. What is the velocity and acceleration of the ball at $t = 6$ s ? (Take $g = 10\ \text{m/s}^2$)
Answer: (A) $20\sqrt{2}$ m/s, $10\ \text{m/s}^2$
Concept. Once released, the ball keeps the car's horizontal velocity and only gravity acts on it.
Step 1. Velocity at release. $5 \times 4 = 20$ m/s horizontally.
Step 2. At $t = 6$ s (2 s later). $v_x = 20$ m/s, $v_y = 10 \times 2 = 20$ m/s:
$$v = \sqrt{20^2 + 20^2} = 20\sqrt2\ \text{m/s}$$
Step 3. Acceleration. $g = 10\ \text{m/s}^2$ downward.
Answer. $20\sqrt2$ m/s, $10\ \text{m/s}^2$ (Option A).
Solution by Sreeraj P, M.Sc Physics