Question

a stone dropped from a 160m high point, and at the same time another stone is thrown up from the ground with speed 40m/s. at what hight from the ground will they meet..?​

Answer 1

Answer:

4m

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Answer 2

Answer:

1.6seconds after release.

Letting the ground to be the origin for this 1-D kinematics problem, taking upward direction as positive and downward direction as negative and writing displacement for both the balls (assuming g=10 m/(s^2) , t=0 at time of release):-

Ball1 :- x1=80 - 5*(t^2)

Ball2 :- x2=50t - 5*(t^2)

When both will meet x1=x2 :-

==> t = 1.6 seconds

Alternatively:-

Using concepts of relative motion, relative speed of approach = 50m/s , relative acceleration = 0 , relative separation = 80m

==> 80 = 50*t ==> t = 1.6 seconds

Explanation: