Posted by **Awonah** on Friday, September 9, 2011 at 11:26am.

A. You can use any coordinate system you like in order to solve a projectile motion problem. To demonstrate the truth of this statement, consider a ball thrown off the top of a building with a velocity at an angle θ with respect to the horizontal. Let the building be 50.0 m tall, the initial horizontal velocity be 9.00 m/s, and the initial vertical velocity be 12.0 m/s. Choose your coordinates such that the positive y-axis is upward, the x-axis is to the right, and the origin is at the point where the ball is released. (a) With these choices, find the ballâ€™s maximum height above the ground, and the time it takes to reach the maximum height. (b) Repeat your calculations choosing the origin at the base of the building.

- Physics -
**Henry**, Sunday, September 11, 2011 at 12:01pm
a. Vf^2 = Vo^2 + 2gd,

d = 50 + (Vf^2 - Vo^2) / 2g,

d = 50 + (0 - (12)^2) / -19.6,

d = 50 + 7.35 = 57.35m above ground.

Vf = V0 + gt,

t = (Vf - Vo) / g,

t(up) = (0 - 12) / -9.8 = 1.22s.

b. d = (Vf^2 - Vo^2) / 2g,

d = (0 - (12)^2) / -19.6 = 7.35m above

ground.

t(up) = (Vf - Vo) / g,

t(up) = (0 - 12) / -9.8 = 1.22s.

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