Posted by **Anonymous** on Saturday, March 15, 2008 at 7:42pm.

A ball is thrown from the top edge of a building with initial velocity components of 15m/s(up) vertically and 20m/s horizontally. It strikes ground 140m from the base of the building. What is the height of the building?

I tried using V2^2 = V1^2 + 2ad and it gave me 11.25 which I know is wrong. How would I calculate this?

- Physics -
**Quidditch**, Saturday, March 15, 2008 at 10:51pm
The ball lands 140m from the base of the building. The balls initial horizontal velocity is 20m/s.

140m = (20m/s) * t

Solve for t.

Since the time the ball is in flight is now known.

Plug that into the "complete" formula for distance:

d = (1/2)*(g)*(t^2) + v*t + h

Let distance 0 be ground level.

initial vertical velocity is v=15m/s

h is the initial distance above the ground which is the height of the building.

0 = (1/2)*(g)*(t^2) + v*t + h

Solve for h.

Remember that the initial velocity is up and the acceleration of gravity is down (watch the signs)

- Physics -
**~christina~**, Saturday, March 15, 2008 at 11:40pm
I helped you with this question yesterday and gave the exact same info as Quidditch Why didn't you use the info I gave?

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