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April 16, 2014

April 16, 2014

Posted by **Ray Dunnigan** on Sunday, June 10, 2012 at 1:36pm.

Pt. A) How far apart are the stones when one has reached a speed of 13.2 m/s^2?

- physics -
**Elena**, Sunday, June 10, 2012 at 2:58pmThe second stone reached the velocity 13.2 m/s for the time

t2=v/g=13.2/9.8 =1.35 s.

The distance covered by the second ball for this time is

h2= =g•t2²/2 =9.8•(1.35) ²/2 =8.9 m

The first stone moved during

t=t1+t2 =1.74+1.35 =3.09 s.

The distance covered by the first stone for 3.09 s is

h1=g•t1²/2 =9.8•(3.09) ²/2 =46.8 m.

Δh=h1-h2= 46.8-8.9= 37.9 m.

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