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November 23, 2014

Posted by **HENRY!** on Wednesday, December 5, 2012 at 2:10pm.

I need help. This is due next week. Please help

- Phyiscs -
**Elena**, Wednesday, December 5, 2012 at 3:28pmThe 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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