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A particle is dropped from height 100m and another particle is projected vertically upward with velocity 50m/s from the ground along the same line.find out the position where two particle will meet
I'm doing this ques with the concept of relative motionLet us consider dropped particle A and upward thrown particle B.(suppose u are a insect and sitting on B , now B will be at rest for u)
Now ,Vab = Va-Vb (velocity of A w.r.t B) (while sitting on A seeing B is Vab)Vab = 0-50 = -50m/s aab= -g-(-g) =0 ( acceleration of A w.r.t B)Distance=100m (as time and distance between 2 particle are independent of frame) Now , since a = 0 it is a simple non-accelerated motion. Applying ,Speed=distance/time 50=100/t=>time=2 sec Now, we can calculate distance travelled by A in time=2sec , that is at which height they will meet.) =>S=ut+at2/2 ( 2nd equation of motion)=>S=(0)(2)-(10)(2)2/2=>S=-20m ( negative sign signifies direction) So, the particles will meet 20m from top or 30m above ground. Here is ur requested answer, there is also a suggestion if u understood just do it without seeing above
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