Question

A car starts from rest and accelerates at $5\, m / s ^{2}$ 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 $\left. g =10\, m / s ^{2}\right)$

• A. $20\, m / s , 5\, m / s ^{2}$
• B. $20\, m / s , 0$
• C. $20\, \sqrt{2}\, m / s , 0$
• D. $20\, \sqrt{2}\, m / s , 10\, m / s ^{2}$

Answer 1

Answer: (D)

Velocity of the car at $t =4 \sec$ is
$V _{ x }= at =4 \times 5=20 \,m / s$
So horizontal velocity $=20 \,m / s$ (remain constant)
Vertical velocity at $t =6 \sec$
i.e. after $2 sec$ of free fall
$V_{y}=g t=20\, m / s$
So net velocity $=\sqrt{20^{2}+20^{2}}=20 \sqrt{2} \,m / s$
and once it starts falling acceleration is only ' $g$ '
i.e., $10 \,m / s ^{2}$