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

You are standing at the foot of the Bank of America building in San Francisco, which is 52 floors (237 m) high. You launch a ball straight up in the air from the edge of the foot of the building. The initial vertical speed is 70 m/s.

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 6:21pm
a. 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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