Question

An elevator at rest which is at the ${10^{th}}$ floor of a building is having a plane mirror fixed on its floor. A particle is projected with a speed $\sqrt 2 m/s$ and at ${45^ \circ }$ with the horizontal. At the very instant of projection, the cable of the elevator breaks and starts falling freely. What will be the separation between the particle and image, $0.5\sec $ after an instant of projection?

Answer 1

Hint: An object that is falling freely under the influence of gravity, it is known as a freely falling object. The objects falling freely under gravity are not affected by the force of air resistance. If the particle is travelling in a downward direction and its speed increases then it is said to be falling freely.

Complete step by step answer:
Given that the particle has a speed of $\sqrt 2 $ and makes an angle of ${45^ \circ }$ with the horizontal, then the velocity of the particle will be resolved into two components.

The horizontal or x-component of the velocity is given by
 ${V_x} = v\cos \theta = \sqrt 2 \cos {45^ \circ }$
$\Rightarrow {V_x} = \sqrt 2 \times \dfrac{1}{{\sqrt 2 }}$
$\Rightarrow {V_x} = 1m/s$
Similarly the vertical or y component of the velocity is given by
$\Rightarrow {V_y} = v\sin \theta = \sqrt 2 \sin {45^ \circ }$
\[\Rightarrow {V_y} = \sqrt 2 \times \dfrac{1}{{\sqrt 2 }}\]
$\Rightarrow {V_y} = 1m/s$
Given that the elevator breaks and the elevator and the ball will now start falling freely, therefore their
acceleration will be due to gravity.
Acceleration of elevator = $g$
Acceleration of ball = $g$
Therefore the acceleration of the particle with respect to the elevator will be zero.
i.e. ${a_{elevator}} = 0$
Now since the separation between the particle and image is to be calculated, therefore using equation of motion
$S = ut + \dfrac{1}{2}a{t^2}$
Since $a = 0$
Substituting the value of acceleration, the equation becomes
$\Rightarrow S = ut$
It is already calculated that the velocity of the particle is $1m/s$ and the given time is $0.5\sec$.
Substituting all the values,
$\Rightarrow S = 1 \times 0.5 = 0.5$
Therefore, the distance between the particle and the mirror is, $u = 0.5m$

The image will also be formed at the same distance as the object is placed, therefore the image distance will be $v = 0.5 + 0.5 = 1m$

Note:
It is to be noted that the air resistance is the force acting on a freely falling object, that has a magnitude equal to the weight of the object and that tends to decrease the velocity of the object. This force depends on the square of the velocity of the object. So when the body accelerates, the velocity increases, and the air resistance also increases. When this force becomes equal to the weight of the object, the object falls with a constant velocity and hence acceleration becomes zero.