A truck starts from rest and accelerates uniformly at 2.0 ms-2. At t = 10 s, a stone is dropped by a person standing on the top of the truck (6 m high from the ground). What are the

(a) velocity, and

(b) acceleration of the stone at t = 11s? (Neglect air resistance.)

Asked by Pragya Singh | 1 year ago |  81

Solution :-

(a)

Initial velocity, u = 0

Acceleration, a = 2 ms-2,

t=10 s

Using equation, v = u + at, we get

v = 0 + 2 x 10 = 20 m/s

The final velocity, v = 20 m/s

At time, t = 11 sec, the horizontal component of the velocity in the absence of the air resistance remains unchanged

Vx =20 m/s

The vertical component of the velocity is given by the equation

Vy = u + ayt

Here t = 11 – 10 = 1s and ay= a = 10 m/s

Therefore, the resultant velocity v of the stone is

$$v=( vx^2 + vy^2) ^ \dfrac{1}{2}$$

v = (20 + 10 2)

v = 22.36 m/s

tan θ = $$\dfrac{v_y}{v_x}$$ = $$\dfrac{10}{20}$$$$\dfrac{1}{2}$$ = 0.5

θ = tan -1 (0.5 ) = 26. 56from the horizontal

(b) When the stone is dropped from the truck, the horizontal force on the stone is zero. The only acceleration of the stone is that due to gravity which is equal to 10 m/s2 and it acts vertically downwards.

Answered by Abhisek | 1 year ago

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