Given,

The initial velocity of the truck, u= 0

Acceleration of the truck,‘a’ = 2 m/s²

Time, t=10 s

According to the first equation of motion, the final velocity, ‘v’ is

v= u+ at

⇒ v= 0 + 2 m/s²× 10 s =20 m/s

The final velocity after t =10s is 20 m/s

(a) At t = 11 s,

The horizontal component of the velocity remains the same, in the absence of air resistance,

Thus, v_x = 20 m/s

According to the first equation of motion, The vertical component of velocity of the stone is given by,

v_y = u + a_yδt

where,

δt = 11 s -10 s = 1 s and

since, the direction is vertical the acceleration acting on it is due to the gravity.

Thus a_y= g =10 m/s²

⇒ v_{y =} 0+ 10 m/s² × 1 s = 10 m/s

The final resultant velocity of the stone is given as,

v_res = (v_x² + v_y²)^1/2

⇒ v_res = (20² + 10²)^1/2 =

→ v_res =22.36 m/s

We suppose that the angle made by the resultant velocity with the horizontal velocity, v_x is θ,

Thus,

tan θ = (v_y/v_x)

⇒ θ = tan^-1(10/20)

⇒ θ = tan^-1(0.5) = 26.57°

The velocity of the stone at t=11 s is 22.36 m/s and is at angle 26.57° with the horizontal

(b) When the stone is dropped from the truck, the horizontal force provided by the truck acting on the stone becomes zero. The only force and thus, the acceleration, that remains is that in the vertical direction i.e. acceleration due to gravity.

Hence, the acceleration of the stone is 10 m/s² and it is in the downward direction.