Physics
posted by Anonymous .
A ball is thrown straight up from the ground with speed v_0. At the same instant, a second ball is dropped from rest from a height H, directly above the point where the first ball was thrown upward. There is no air resistance.
Find the value of H in terms of v_0 and g so that at the instant when the balls collide, the first ball is at the highest point of its motion.

The highest point is when vertical velocity is zero.
Vf^2=Vi^2+2gh
h= vi^2/2g= Vi^2/19.8
Now, you need for the top ball to fall Hh in the same time.
time= distance/avgvelocityfor the bottom ball
= Vi^2/2g*Vi/2= Vi/g
now to the top:
Hh=1/2 g t^2= 1/2 g (Vi^2/g^2)=1/2 Vi/g
Now solve for H.
check all that.
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