a stone is dropped from a height 300m and at the same time another stone is projected vertically upwards with a velocity 100m and find the time when the two stones meet
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Consider the motion of the ball dropped from the height.
Let the balls meet after time t and displacement from the above x.
We have, s=ut+(1/2)at2
x=(1/2)gt2=4.9t2
x=4.9t2
Consider the motion of second ball, thrown vertically above.
s=ut+(1/2)at2
-(300-x) = -100t +(1/2)gt2
x-300= -100t+4.9t2
x-300=-100t+x
100t=300t
t=3s
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