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Question: A car starts from rest with a constant acceleration of \(2\;{\rm{m/}}{{\rm{s}}^2}\). After \(5\;{\rm...

A car starts from rest with a constant acceleration of 2  m/s22\;{\rm{m/}}{{\rm{s}}^2}. After 5  s5\;{\rm{s}} a ball is dropped through the window of the car. The window of the car is at a height of 1.5  m1.5\;{\rm{m}} from the ground. What will be the speed of the ball 0.5  s0.5\;{\rm{s}} after it was dropped? (Take g=10  m/s2g = 10\;{\rm{m/}}{{\rm{s}}^2}).

  1. 5  m/s5\;{\rm{m/s}}
  2. 5  m/s\sqrt 5 \;{\rm{m/s}}
  3. 55  m/s5\sqrt 5 \;{\rm{m/s}}
  4. 2  m/s2\;{\rm{m/s}}
Explanation

Solution

Velocity is a physical quantity that is vector and measures displacement over the change in time. If we neglect air resistance, all objects, regardless of their mass, fall to earth with the same acceleration.

Complete step by step answer:
Since the car is starting from the rest, we have the initial velocity, u=0u = 0. So we can here apply the formula to calculate velocity. As we know, velocity is the product of acceleration and time. This can be expressed as, v=u+atv = u + at. So as the given acceleration is 2  m/s22\;{\rm{m/}}{{\rm{s}}^2} and time is 5  s5\;{\rm{s}}, we can calculate the velocity as, v=0+5  s×2  m/s2=10  m/sv = 0 + 5\;{\rm{s}} \times 2\;{\rm{m/}}{{\rm{s}}^2} = 10\;{\rm{m/s}}. We know that if a body is dropped from an accelerated system, the force acting on it no longer acts. Therefore, when the stone is dropped out of the car at 5  s5\;{\rm{s}}, the stone will not possess any acceleration.

Therefore after 5  s5\;{\rm{s}}, the stone possesses the following two motions:- uniform motion with velocity 10  m/s10\;{\rm{m/s}}along the horizontal, and accelerated motion under gravity. Therefore after 0.5  s0.5\;{\rm{s}} after it was dropped, the time is t=5.5  st = 5.5\;{\rm{s}}. At this time, we have the velocity of the stone along horizontal as, vx=10  m/s{v_x} = 10\;{\rm{m/s}}. And the velocity of the stone along the vertical at t=5.5  st = 5.5\;{\rm{s}}, that is, 0.5  s0.5\;{\rm{s}} after being dropped out of the car is
v=0+gt ⇒v=0+10×0.5 ⇒v=5  m/s\begin{array}{c} v = 0 + gt\\\ \Rightarrow v= 0 + 10 \times 0.5\\\ \Rightarrow v= 5\;{\rm{m/s}} \end{array}

So, the correct answer is “Option A”.

Note:
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/s210\;{\rm{m/}}{{\rm{s}}^2}and it acts vertically downward. A ball falling under the influence of gravity is an example of what we call motion with constant acceleration.