Correct option is (4) 20√2 m/s, 10m/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 = \(\sqrt{v_h^2 + v_v^2}\)
\(= \sqrt{20^2+20^2}\)
\(= 20\sqrt 2\) m/s
The motion of the ball is under gravity, so,
anet = 10 m/s2