Two particles A and B are projected simultaneously from two points on a horizontal ground as shown

Answer

Verified

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-axisHeight 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.