A car starts from rest and accelerates uniformly with 2 m/s². At t = 5s, a stone is dropped out of the window 15m high of the car. What is velocity and acceleration of stone at t = 5.5 s? (g = 10m/s²)

Given the car starts from rest.
That is initial velocity, u = 0 m/s
Acceleration of the car = 2 m/s².
Therefore,
Velocity of car at t = 5s is 10 m/s.
When the stone is dropped, the velocity of the car and hence velocity of stone is 10 m/s in the horizontal direction.
At t= 5s the stone loses the contact from the car. Therefore, the stone will move uniformly in the horizontal direction and accelerate in vertically downward direction due to gravity.

After 0.5 seconds, the stone is dropped. 
On resolving velocity, we get X and Y as it's components. 
V_x = v_x + a_xt = 10 + 0 X 0.5 = 10 m/s
V_y = v_y + a_yt ₌ 0 + 10 X 0.5 = 5 m/s
Acceleration of stone at t= 5 sec is the only acceleration due to gravity. 
That is, a = 10 m/s².