posted by Kelsey .
A baseball is thrown from the roof of a 22.3 m tall building with an initial velocity of 13.1 m/s and is directed 50.9 degrees above the horizontal.
A)What is the speed of the ball just before it strikes the ground?
B)What is the answer for part A) if the initial velocity was directed at an angle of 50.9 degrees below the horizontal?
A) The horizontal velocity component will remain
Vx = 13.1 cos 50.9
and the vertical component will increase so that
Vy^2 - (13.1 sin 50.9)^2 = 2 g H
where H is the height of the building.
B) You will get the same answer, since the (13.1 sin theta)^2 term is squared in the Vy equation, and cos theta = cos(-theta)