Two particles $ A $ and $ B $ are projected simultaneously from ground towards each other as shown. If they collide in mid-air then, the height above the ground where they collide is:

(A) $ 75m $
(B) $ 25m $
(C) $ 100m $
(D) $ 125m $

Hint: We will solve this question with the help of basic equations of projectile motion and relative motion. Projectile motion is a form of motion experienced by an object or particle that is projected near the Earth's surface and moves along a curved path under the action of gravity only.
Formula Used:
The formula for height of the particle at any time $ t $ sec:
 $ h = ut + \dfrac{1}{2}a{t^2} $
Where
 $ t $ is time in seconds
 $ h $ is height of the particle at any point of time
 $ u $ is the initial speed of the particle
 $ a $ is the acceleration due to gravity

Complete Step-by-Step Solution:
Let us suppose that both the particles meet at a height $ h $ from ground level.
Then in X-axis,
Distance between both the particles is $ 140m $
 $ {v_a} $ in X-axis $ = 100\cos {53^\circ } $
 $ = 100 \times \dfrac{3}{5} = 60m/s $
The relative velocity between the two particles is given by
 $ {v_{ab}} = {v_a} - {v_b} $
 $ \Rightarrow {v_{ab}} = 60 - ( - \dfrac{{{v_b}}}{{\sqrt 2 }}) $
Hence we get,
 $ \Rightarrow {v_{ab}} = 60 + \dfrac{{{v_b}}}{{\sqrt 2 }} $
So, the time of collision, $ t = \dfrac{{140}}{{60 + \dfrac{{{v_b}}}{{\sqrt 2 }}}} $
In Y-axis
Height at which they collide is same as $ h $
 $ h = {v_a}\sin {53^\circ } - \dfrac{1}{2}g{t^2} $ …………………. (i)
Also,
 $ h = {v_b}\sin {45^\circ } - \dfrac{1}{2}g{t^2} $ …………………(ii)
 $ \Rightarrow 100\sin {53^\circ } + ( - \dfrac{1}{2} \times 10 \times {t^2}) = {v_b}\sin 45 \times t - \dfrac{1}{2}g{t^2} $
Hence we get,
 $ \Rightarrow 100 \times \dfrac{4}{5} = \dfrac{{{v_b}}}{{\sqrt 2 }} $
 $ \therefore {v_b} = 80\sqrt 2 m/s $
Time of collision $ = \dfrac{{140}}{{60 + \dfrac{{80\sqrt 2 }}{{\sqrt 2 }}}} $ $ = 1\sec $
Now we put the values of $ t $ in equation (i), we get
 $ h = 80 + ( - \dfrac{1}{2} \times 10 \times {t^2}) $
 $ \Rightarrow h = 80 \times 1 - \dfrac{1}{2} \times 10 \times {1^2} $
Therefore we get,
 $ \therefore h = 75cm $
Now we will conform our answer by putting the value in equation (ii)
 $ \Rightarrow h = 80\sqrt 2 \times \dfrac{1}{{\sqrt 2 }} \times 1 - \dfrac{1}{2} \times 10 \times {1^2} $
 $ \Rightarrow h = 80 - 5 = 75m $

Hence the correct answer is option A.

Note: We should always confirm the answer by putting the value of $ t $ in the equation (i) and (ii) obtained. This will correct our errors in case we have accidently made and our final answer will be more accurate.