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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

Asif , 7 Years ago
Grade 11
anser 1 Answers
Arun

Last Activity: 7 Years ago

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 (vx) of velocity, in the absence of air resistance, remains unchanged, i.e.,
vx = 20 m/s
The vertical component (vy) of velocity of the stone is given by the first equation of motion as:
vy = u + ayδt
Where, δt = 11 – 10 = 1 s and ay = g = 10 m/s2
∴vy = 0 + 10 × 1 = 10 m/s
The resultant velocity (v) of the stone is given as:

v = (vx2 + vy2)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, vx
∴ tan θ = (vy / vx)
θ = tan-1 (10 / 20)
= 26.570

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