Question

You shoot an arrow straight ahead; it leaves your bow moving parallel to the ground. You find it embedded in the ground 50m away from where you shot it. It makes an 82 degree angle with the level ground , reflecting the direction it was traveling at the time it hit. How fast was it going when it left your bow? (You can neglect air resistance)

So I figured this out...

yf=o
yf=yi+vyi(t)+1/2gt^2
vyi=0

vf= gt
so -yi=1/2gt^2


and xf=vx/t
vx remains constant

Fnet=Fg=-mg since gravity is the only thing acting on the arrow

<cos82, cos8,cos90> = unit vector of the arrow based on how it landed on the ground

unit vector = <vx,-gt,0>/vector magnitude

also
unit vector=<50m/t,-gt,0>vector magnitude

But then I got stuck, where should I go from here if I don't know the time or magnitude of the vector?

Answer 1

well, if fell a vertical distance of h in time t.
h=.5gt^2 or t=sqrth/2g
vv=g t or v=sqrt hg/2

but the horizontal velocity was 50/t

now looking at the angle at the end..
tan82=horizonal/vertical=50/t / gt
tan82=50/gt^2
solve for time t in the air.
then velocity initiual= 50/t