Projectile Question involving angular velocity and acceleration

[m84uily]

Homework Statement

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 57.0m away, making a 2.00° angle with the ground.

vix = ?m/s
vfx = vix
viy = 0
vfy = ?m/s
dx = 57.0m
dy = ?
ax = 0
ay = -9.8m/s^2
dΘ = -2.00°
t = ?

Homework Equations

d = (vi)(t) + (1/2)(a)(t^2)
vf^2 = (vi^2) + (2)(a)(d)
vf = vi + (a)(t)
d = ((vi + vf)/2)(t)

The Attempt at a Solution

dΘ = -2.00°
ay = -9.8m/s^2

ω = angular velocity
α = angular acceleration

dΘ = (ωi)(t) + (1/2)(α)(t^2)

dx = (vix)(t) + (1/2)(ax)(t^2)
dx = (vix)(t)
(dx)/(vix) = t

dΘ = (ωi)(dx)/(vix) + (1/2)(α)(((dx)/(vix))^2)
dΘ = (1/2)(α)(((dx)/(vix))^2)

Well, this is as far as I know how to take it.

Any help is appreciated! Thanks in advance.

 

[tiny-tim]

welcome to pf!

hi m84uily! welcome to pf!

(try using the X² icon just above the Reply box )

no, this has nothing to do with angular velocity or angular acceleration

the only relevance of the 2° angle is that it tells you the direction of v_f

just use the standard equations for constant acceleration for the y direction, and for zero acceleration for the x direction

 

[m84uily]

Thanks! I'll try that!

 

[m84uily]

Sorry but I'm still unsure of how to solve this!

I now understand that ^{arctan(|vfy|/|vfx|) = 2°}

But I don't know how to calculate either vfy or vfx, here's an attempt.


^{(vfy)^2 = (viy^2) + 2(ay)(dy)
vfy = ((-19.6)(dy))^(1/2)}


So, time is needed in order to solve for dy.And time is also necessary for vfx.

I'm not sure what to do! More help would be appreciated.

EDIT: I got it! Sorry I just panicked and assumed I couldn't do it.