Question

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 

1. velocity, and 
2. acceleration of the stone at t = 11s ? 
   (Neglect air resistance.)

Answer 1

(a) 22.36 \(m/s\), at an angle of 26.57° with the motion of the truck.

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(b) 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 × 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 (\(\text v_x\)) of velocity, in the absence of air resistance, remains unchanged, i.e., 
\(\text v_x\) = 20 \(m/s\)
The vertical component (\(\text v_y\)) of velocity of the stone is given by the first equation of motion as: \(\text v_y\) = \(u+ayt\)
Where, t = 11 – 10 = 1 s and ay = g = 10 \(m/s^2\)
\(\text v_y\)= 0 + 10 × 1 = 10 \(m/s\)
The resultant velocity (v) of the stone is given as:

v = \(\sqrt{v_x^2+v_y^2}\)
= \(\sqrt{20^2+10^2}\)= \(\sqrt{400+100}\)
= \(\sqrt{500}\) = 22.36 \(m/s\)
Let \(\theta\) be the angle made by the resultant velocity with the horizontal component of velocity, \(\text v_x\)
tan \(\theta\) = \(\bigg(\frac{v_y}{v_x}\bigg)\)
\(\theta\) = \(\tan^{-1}\)\(\bigg(\frac{10}{20}\bigg)\)
= \(\tan^{-1}\)(0.5)
= 26.57°
When the stone is dropped from the truck, the horizontal force acting on it becomes zero. However, the stone continues to move under the influence of gravity. 

Hence, the acceleration of the stone is 10 \(m/s^2\) and it acts vertically downward.