Physics
posted by Anonymous .
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?

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) 
I helped you with this question yesterday and gave the exact same info as Quidditch Why didn't you use the info I gave?

Thanks! I saw it and used it but I'm not sure if I did it right. I got 140 for the height of the building and I don't think that's right.

I got something in that range, but I believe your answer has more error than I would expect.
What did you get for time that the ball was in the air?
What numbers did you plug into your final equation? 
For time t = d/v
140/20
t= 7
0 = 1/2(10)(7^2)+(15)(7)+h
h = 140
h = 140
We were asked to use 10 for our acceleration. 
You got it!
Using 10m/(s^2) for gravity, your answer is correct.
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