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?