Posted by James on Wednesday, April 25, 2012 at 1:07am.
A ball is thrown from a building standing 30m high, landing 10m away from the building 3.2 seconds after it was thrown. Find the angle above the horizontal and the initial velocity of the ball

Physics  Damon, Wednesday, April 25, 2012 at 1:42am
0 = 30 + Vi (3.2)  4.9 (3.2)^2
Vi = 6.31 up
10 = u (3.2)
u = 3.13
tan theta = 6.31/3.13
theta = 63.6 degrees up from horizontal
speed = sqrt(3.13^2 + 6.31^2)
= 7.04