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.

 

[rwisz]

I would approach this problem by first defining the variables and equations that are relevant to the situation. In this case, we are dealing with projectile motion, which involves both linear and angular motion. The variables given in the problem are vix, vfx, viy, vfy, dx, dy, ax, ay, dΘ, and t. The equations given are for linear motion, but we also need to consider the equations for angular motion.

For linear motion, we can use the equations d = (vi)(t) + (1/2)(a)(t^2) and vf = vi + (a)(t) to solve for the unknown variables vix and vfy. We know that viy = 0, so we can use this to solve for vfy. We also know that dx = 57.0m, so we can use this to solve for t. Once we have vix and vfy, we can use the equation vf^2 = (vi^2) + (2)(a)(d) to solve for vfx.

For angular motion, we can use the equation dΘ = (ωi)(t) + (1/2)(α)(t^2) to solve for the unknown variable ωi. We also know that dΘ = -2.00°, so we can use this to solve for ωi. Once we have ωi, we can use the equation ωf = ωi + (α)(t) to solve for α.

Finally, we can use the equations d = ((vi + vf)/2)(t) and dΘ = (1/2)(α)(((dx)/(vix))^2) to solve for dy. We know that dx = 57.0m and dΘ = -2.00°, so we can use these values along with the values we have already calculated to solve for dy.

In summary, to find the initial angular velocity (ωi) and angular acceleration (α), we can use the equations dΘ = (ωi)(t) + (1/2)(α)(t^2) and ωf = ωi + (α)(t). To find the initial linear velocity (vix) and final linear velocity (vfx), we can use the equations d = (vi)(t) + (1/2