Posted by Anonymous on Thursday, March 13, 2008 at 7:55pm.
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  ~christina~, Thursday, March 13, 2008 at 9:32pm
since you have the horizontal or x velocity you can calculate the time needed for the ball to reach the x distance(140m)
using this equation:
xfxi= v_{i}+ 1/2at^2
the thing is there is no acceleration in x direction so the a is 0 thus that part cancels out of the equation to make:xfxi= v_{i}
Then use this equation, find t for the time.
After this, use t and then plug it into the equation again (same one) However now that you have t it took to reach the ground in x direction it is the (same in the y) direction
you can use the same equation but this time there IS a acceleration in the y direction.
yfyi= v_{i}+ 1/2at^2
so use the same equation but now plug in initial v for the vertical or y direction and t that you found before and also plug in 9.8m/s^2 for the acceleration. And you should find your answer.
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