A car starts from rest and accelerates at 5 m/s2. 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?
 A car starts from rest and accelerates at 5 m/s2. 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?

(1) 20 m/s, 5 m/s2

(2) 20 m/s, 0

(3) 20 2 m/s, 0

(4) 20 2 m/s, 10 m/s2

Solution: Answer (4)

Initial velocity of car = 0

Acceleration of car = 5 m/s2

Velocity of car at t = 4 s;

v = u + at

Therefore, v = 0 + 5 × 4 = 20 m/s

At t = 4 s,

A ball is dropped out of a window so the velocity of the ball at this instant is 20 m/s along horizontal.

After 2 seconds of motion :

Horizontal velocity of ball = 20 m/s (∵ ax = 0)

Vertical velocity of ball (vy) = uy + ayt

Vy = 0 + 10 × 2 = 20 m/s (∵ ay = g = 10 m/s2)

So magnitude of velocity of ball

$v=\sqrt{v_{x}^{2}+v_{y}^{2}}=20\sqrt{2}m/s$

Acceleration of ball at t = 6 s is g = 10 m/s2

As ball is under free fall