A truck starts from rest and accelerate uniformly with 2m per second square at t=10s a stone is dropped by a person standing on the top of truck 6m high from the ground what is the velocity at t=11s take g=10

Initial velocity of the truck, u = 0
Acceleration, a = 2 m/s²
Time, t = 10 s
As per the first equation of motion, final velocity is given as:
v = u + at
= 0 + 2 × 10 = 20 m/s
The final velocity of the truck and hence, of the stone is 20 m/s.
At t = 11 s, the horizontal component (v_x) of velocity, in the absence of air resistance, remains unchanged, i.e.,
v_x = 20 m/s
The vertical component (v_y) of velocity of the stone is given by the first equation of motion as:
v_y = u + a_yδt
Where, δt = 11 – 10 = 1 s and a_y = g = 10 m/s²
∴v_y = 0 + 10 × 1 = 10 m/s
The resultant velocity (v) of the stone is given as:

v = (v_x² + v_y²)^1/2
= (20² + 10²)^1/2
= 22.36 m/s

Let θ be the angle made by the resultant velocity with the horizontal component of velocity, v_x
∴ tan θ = (v_y / v_x)
θ = tan^-1 (10 / 20)
= 26.57⁰