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April 18, 2014

April 18, 2014

Posted by **Sara** on Sunday, August 26, 2012 at 5:37pm.

a. How high up does the ball go?

b. How fast is the ball going right before it hits the top of the building?

c. For how many seconds total is the ball in the air?

(Not sure which equation to use, that's the main reason why I'm asking the question. I don't want to use the wrong equation.)

- Physics -
**drwls**, Sunday, August 26, 2012 at 6:21pma. Use conservation of energy. The ball rises to a maximum height H such that

gH = Vo^2/2

H = (1/g)*(1/2)*Vo^2 = 250 m

Vo is the initial velocity that it is thrown.

b. At the top of the building, where Y = 237 m,

Vo^2/2 = V^2/2 + g*Y

Solve for V

V^2/2 = (1/2)(4900)-(g*237)

= 2450 - 2323 = 127 m^2/s^2

V = 15.9 m/s

c. It takes Vo/g = 7.14 s to go up and the same time to come down.

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