You shoot an arrow into the air. 2.oo seconds later the arrow has gone straight upward to a height of 30 meters above its launch point. (a) What was the arrow's initial speed? (b) How long did it take for the arrow to first reach a height of 15 meters above it's launch point?

Answer 1

Y = Vo*t -(g/2)t^2
At t = 2.00s,
30 = 2*Vo -(4.9)*4
(a) Vo = 24.8 m/s

(b) Now that you know Vo, use the first equation again to solve for t when Y = 15 m.

Answer 2

3.45

Answer 3

5.6