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A stone is dropped from topp of a tower 300m high and at the same time another stone is projected vertically upwards with a velocity of 100m/s from the

foot of the tower. find when and where the two stones meet.


Upward motion-
u = 100m/s g= -10
s=ut + 1/2 gt^2
s= 100t - 1/2 (10)t^2
s= 100t - 5t^2........1
downward motion
300- s = 100t + 5t^2.........2
adding (1) & (2)
from eqn. 1
s = 100*3/2-5*(3/2)^2
and answer will give u the height at which the stones will meet from the ground
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