The correct option is A 20√2 m/s,10 m/s2
Velocity of the car at t=4 s,
v=u+at=0+5×4=20 m/s
The ball is dropped at this instant, this means the horizontal velocity of the ball will be vh=20 m/s.
This velocity will remain constant as there is no horizontal acceleration.
Now, vertical velocity at t=6 s i.e. Δt=2 s,
vv=uv+avΔt=0+10×2=20 m/s.
Further, the net velocity is,
vnet=√v2h+v2v=√202+202=20√2 m/s.
The motion of the ball is under gravity, so,
anet=10 m/s2
Hence, option (A) is the correct answer.