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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
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

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3 years ago

```							Initial velocity of the truck, u = 0Acceleration, a = 2 m/s2Time, t = 10 sAs per the first equation of motion, final velocity is given as:v = u + at= 0 + 2 × 10 = 20 m/sThe 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/sThe vertical component (vy) of velocity of the stone is given by the first equation of motion as:vy = u + ayδtWhere, δt = 11 – 10 = 1 s and ay = g = 10 m/s2∴vy = 0 + 10 × 1 = 10 m/sThe resultant velocity (v) of the stone is given as:v = (vx2 + vy2)1/2= (202 + 102)1/2= 22.36 m/sLet 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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3 years ago
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