Question

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?

Answer 1

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.