college physics
posted by tina
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

tina
(15.0)(1/2)(9.80)(.96)/.96 answer equals .10.7

Damon
Tina  I did this problem for Brittany, scroll down.

Damon
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

Now how long did it take to reach 3.1 m down
3.1 = 4.9t^2
t = .795 seconds

so the second rock spends 1.75 .795 or .955 seconds in the air
15 = Vo (.955) 4.9 (.955)^2
.955 Vo = 3.6415
Vo =  11.9
so 11.9 m/s downward (negative) 
Camrik
wrong answer 4.9(.955)^2 does not equal 3.64
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