Asked 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?
Answered by ~christina~
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:
xf-xi= v<sub>i</sub>+ 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:xf-xi= v<sub>i</sub>

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.

yf-yi= v<sub>i</sub>+ 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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