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Solutions
Initial velocity of car = 0
Acceleration of car = 5 m/s2
Velocity of car at t = 4 s; v = u + at
⇒ v = 0 + 5 × 4 = 20 ms–1At t = 4 s, A ball is dropped out of a window so velocity of ball at this instant is 20 ms–1 along horizontal.
After 2 seconds of motion :
Horizontal velocity of ball = 20ms−1(∵ ax = 0)
Vertical velocity of ball (vy) = uy + ayt
vy = 0 + 10 × 2 = 20ms−1 (∵ ay = g = 10m∕s2)
So magnitude of velocity of ball

Acceleration of ball at t = 6s is g = 10m∕s2
As ball is under free fall.