Question

Us, a
stone is dropped by a person standing on the top of the truck (6 m high from
A truck starts from rest and accelerates
the ground). What are the (a) velocity, and (b) acceleration of the stone at in
11s? (Neglect air resistance.)
no from the ceiling of a room by a string 2 m long in​

Answer 1

Answer: (a) Initial velocity of the truck, u = 0

Acceleration, a = 2 m/s  

2

 

Time, t = 10 s

As per the first equation of motion, final velocity is given as:

v = u + at

= 0 + 2 x 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 + ayt

Where, t = 11 - 10 = 1 s and ay = g = 10 m/s  

2

 

vy = 0 + 10 x 1 = 10 m/s

The resultant velocity (v) of the stone is given as:

v = (vx  

2

 + vy  

2

)  

1/2

 

= (20  

2

 + 10  

2

)  

1/2

 

= 22.36 m/s

Let  be the angle made by the resultant velocity with the horizontal component of velocity, vx be θ

tanθ=vy/vx

θ=tan  

−1

(vy/vx)

= tan  

−1

 (10 / 20)

= 26.570

Explanation: hope this helps