You are watching an archery tournament when you start wondering how fast an arrow is shot from the bow. Remembering your physics, you ask one of the archers to shoot an arrow parallel to the ground. You find the arrow stuck in the ground 68.0 m away, making a 3.00o angle with the ground.
How fast was the arrow shot?
The Attempt at a Solution
First a made a triangle and from that I said tan(3)=y/68 and got 3.564 which would be the initial height. Then I used that equation and said 0=3.564+vi*t-4.9t2 I then used Vf=Vi+at so 0=vi-9.8t and vi=9.8t. I plugged that into the other equation which got me 3.564+9.8t2 -4.9t2 I did that out and got t=.853 Then I plugged that into x=Vi*t so 68=Vi*.853 and got vi=79.72 m/s and the online thing says to try again. I do not know where I went wrong. Thanks for any help!