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=v₀-gt, 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=5.5-2.8 = 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 37.2-35.7=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=v₀t₂-gt₂²/2,

therefore, h₁+h₂=1.5 =>

v₁t₂+gt₂²/2 + v₀t₂-gt₂²/2 =1.5

(v₁+v₀)t₂=1.5

t₂=1.5/(v₁+v₀)=1.5/26.5+27=0.028 s.

h₂=v₀t₂-gt₂²/2 =

=27•0.028 -9.8•0.028²/2=0.75 m

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