Posted by Todd on Wednesday, June 30, 2010 at 10:46pm.
You shoot an arrow into the air. Two seconds later (2.00 {\rm s}) the arrow has gone straight upward to a height of 32.0 m above its launch point.
What is the initial velocity and how long did it take for the arrow to reach 16.0 m above the launch point?

Physics  drwls, Wednesday, June 30, 2010 at 11:15pm
Velocity (v) obeys the equation
v = Vo  gt = Vo  9.8 t
where Vo is the initial velocity
At t=2 s
32 = Vo  19.6 m
Therefore Vo = 51.6 m/s
y (the height) = Vo t  (g/2)t^2
= 51.6 t  4.9 t^2
Set y = 16 and solve for the time t when that height is reached. Take the lowest of the two solutions of the quadratic equation; the other is the time coming back down.