The correct option is B 20√2 m/s, 10 m/s2
Velocity of the car at t=4 s, using kinematic equation,
v=u+at
where u=0 ; a=5 m/s2 ; t=4 s
v=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, using kinematic equation,
v=u+at
where v=vv ; u=uv=0 ; a=av=10 m/s2 ; t=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=g=10 m/s2
Hence, option (A) is the correct answer.