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 ball’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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