Posted by HENRY! on Wednesday, December 5, 2012 at 2:10pm.
Suppose you throw a stone straight up with an initial velocity of 27.0 m/s and, 5.5 s later you throw a second stone straight up with the same initial velocity. The first stone going down will meet the second stone going up. At what height above the point of release do the two stones meet?
I need help. This is due next week. Please help

Phyiscs  Elena, Wednesday, December 5, 2012 at 3:28pm
The height of the first stone is
h₀=v₀²/2g=27²/2•9.8 = 37.2 m.
v=v₀gt, v=0 =>
the time of the upward motion is
t= v₀/g =27/9.8=2.8 s.
The 1st stone was falling down during
t₁=5.52.8 = 2.7 s, when the 2nd stone started.
The 1st stone covered the vertical distance
y=gt₁²/2 =9.8•2.7²/2=35.7 m
and it had a speed
v₁=gt₁=9.8•2.7 =26.5 m/s.
Now at the instant of the second stone
start, the distance between
them is 37.235.7=1.5 m.
If the time from the start of the 2nd stone
and the collision is t₂, the distance covered by the 1st stone is
h₁=v₁t₂+gt₂²/2,
and the distance covered by the 2nd stone is
h₂=v₀t₂gt₂²/2,
therefore, h₁+h₂=1.5 =>
v₁t₂+gt₂²/2 + v₀t₂gt₂²/2 =1.5
(v₁+v₀)t₂=1.5
t₂=1.5/(v₁+v₀)=1.5/26.5+27=0.028 s.
h₂=v₀t₂gt₂²/2 =
=27•0.028 9.8•0.028²/2=0.75 m
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