Question

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
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

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