Posted by **tina** on Sunday, January 31, 2010 at 7:58pm.

While standing on a bridge 15.0 m above the ground, you drop a stone from rest. When the stone has fallen 3.10 m, you throw a second stone straight down. What initial velocity must you give the second stone if they are both to reach the ground at the same instant? Take the downward direction to be the negative direction

- college physics -
**tina**, Sunday, January 31, 2010 at 8:01pm
(-15.0)-(1/2)(-9.80)(.96)/.96 answer equals .10.7

- college physics -
**Damon**, Sunday, January 31, 2010 at 8:05pm
Tina - I did this problem for Brittany, scroll down.

- Here Again -
**Damon**, Sunday, January 31, 2010 at 8:07pm
first get time to fall to ground

a = -9.8

v = -9.8 t

-15 = -4.9 t^2

t = sqrt (15/4.9) = 1.75 seconds

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Now how long did it take to reach 3.1 m down

3.1 = -4.9t^2

t = .795 seconds

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so the second rock spends 1.75 -.795 or .955 seconds in the air

-15 = Vo (.955) -4.9 (.955)^2

.955 Vo = 3.64-15

Vo = - 11.9

so 11.9 m/s downward (negative)

- college physics -
**Camrik**, Tuesday, September 6, 2011 at 7:40pm
wrong answer -4.9(.955)^2 does not equal 3.64

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