NEET 2021 Physics and Chemistry

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–1
At 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
(v)=√vx2+vy2=20√2m∕s
Acceleration of ball at t=6s is g=10m∕s2
As ball is under free fall.