Speed of an arrow using distance and angle with the ground

[kerbyjonsonjr]

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 68.0 m away, making a 3.00^o angle with the ground.
How fast was the arrow shot?

Homework Equations

x_f=x_i +vi*t+1/2at²
v_f=v_i +at

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 v_i=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 v_i=79.72 m/s and the online thing says to try again. I do not know where I went wrong. Thanks for any help!

 

[Delphi51]

First a made a triangle and from that I said tan(3)=y/68

This assumes the arrow flew in a straight line, but in fact the path is parabolic.

Consider instead that at the end of the flight tan (3) = Vy/Vx.

 

[RTW69]

You can assume the arrow is shot at ground level at an angle of 3 degrees. In any event I don't see an error in your work. See if the online solution takes 80 m/s.

 

[kerbyjonsonjr]

This assumes the arrow flew in a straight line, but in fact the path is parabolic.

Consider instead that at the end of the flight tan (3) = Vy/Vx.

You are absolutely right. But how would I find Vy or Vx?

 

[mburt]

Given:

d_x = 68.0 m
angle = 3.00^o

And...

v_x = vcos(3)
v_1y = vsin(3)

d_y = v_1yt + 0.5at²
d_y = vsin(3) - 4.9t²

... so it looks like there is three unknowns there.

d_x = vcos(3)t
68.0 m = vcos(3)t
t = 68.0m / vcos(3)

Substitute in first formula:

d_y = vsin(3) - 4.9(68.0 m / vcos(3))²

... and this is as far as I can get! Without a vertical displacement I'm really not sure how this problem could be solved! I'm going to try to work it out on paper now, so I'll see what I get

 

[kerbyjonsonjr]

I'm really not sure how this problem could be solved!

Me either! It is nice to know I am not the only one that thinks that. It is pretty bogus. Let me know if you have any luck.

 

[RTW69]

You can sole for Vo directly.
x=Vo*cos(theta)*t
68=Vo*cos(3)*t
68.093=Vo*t

Y=Vyo*t-.5*g*t^2 but Y=0 at landing
Vyo=.5*g*t=Vo*sin(theta)
t=.0107*Vo

2 equations two unknowns

Vo=80 m/3

 

[kerbyjonsonjr]

You can sole for Vo directly.
x=Vo*cos(theta)*t
68=Vo*cos(3)*t
68.093=Vo*t

Y=Vyo*t-.5*g*t^2 but Y=0 at landing
Vyo=.5*g*t=Vo*sin(theta)
t=.0107*Vo

2 equations two unknowns

Vo=80 m/3

That's what I got. And that seems like a perfectly feasible speed but the dumb online thing says that it is not the correct answer.

 

[mburt]

The displacement is negative? The question says the arrow is released parallel to the ground, I don't see how it could be zero.

For example, if the arrow is released 1.3 m above the ground, then the overall y-displacement is -1.3 m.

 

[mburt]

I've also just realized that v_1y equals 0, since it is shot parallel to the ground. So v₁ = v_x... so basically if you can find v_x, then that's your answer...

Not sure how though, really.

 

[kerbyjonsonjr]

I've also just realized that v_1y equals 0, since it is shot parallel to the ground. So v₁ = v_x... so basically if you can find v_x, then that's your answer...

Not sure how though, really.

I am tempted just to hit the "show answer" button just to see what they give for an answer.

 

[RTW69]

As I think about it x=136 meters not 68 meters. The only place where the velocity of the arrow is in the horizontal direction is at the top of the parabola, which happens in mid-flight. Set x=136 meters, theta =3 degrees. I get Vo=112.8 m/s. Does that work?

 

[kerbyjonsonjr]

As I think about it x=136 meters not 68 meters. The only place where the velocity of the arrow is in the horizontal direction is at the top of the parabola, which happens in mid-flight. Set x=136 meters, theta =3 degrees. I get Vo=112.8 m/s. Does that work?

YOU GOT IT! VERY NICELY DONE! THANK YOU! What equation did you plug those numbers into to get Vo?

 

[RTW69]

see post #7 except x=136 and not 68.

136.2=Vo(.0107Vo)

 

[kerbyjonsonjr]

see post #7 except x=136 and not 68.

136.2=Vo(.0107Vo)

OK Thanks! Why do you double x though? Also, you have Y=Vyo*t-.5*g*t^2 but then you go to Vyo=.5*g*t=Vo*sin(theta) I was wondering what happened to the t squared? How come you don't get t²= .0107*Vo?

 

[RTW69]

136 is twice 68. The arrow is shot horizontal to the ground and has no vertical velocity component. The only place on the flight of a projectile where you only have a horizontal velocity component is at the very top of the parabola. So the arrow is shot essentially at the top of the parabola which represents 1/2 of the total flight distance if the arrow was shot from ground level.

Y=Vyo*t-.5*g*t^2

0=Vyo*t-.5*g*t^2

Vyo*t= .5*g*t^2

Vyo=.5*g*t
But Vyo also = Vo sin(theta)

 

[Delphi51]

Doubling the distance is clever, but it obscures the basic simplicity of the question! All you need is tan(3) = Vy/v (v is initial horizontal speed). Since Vy = 0t + at = gt this is tan(3) = gt/v or
v = gt/tan(3) [1]
Horizontally, x = vt = gt²/tan(3) [2]
Using x = 68 gives t = 0.6027.
Put this in [1] to get v = 112.8
I used g = 9.81