How Fast Does an Arrow Travel in Archery?

[a.d.d92]

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 66.0 m away, making a 3.00 degree angle with the ground.

- How fast was the arrow shot?

My attempt to the solution:
d = 66m
a = g*sin(theta)
= 0.51289 m/s^2
Sf = 66m
Si = 0m

Vf^2 = Vi^2 + 2a(Sf-Si)
Vf^2 = 0 + 2(0.512)(66-0)
Vf = 8.22 m/s

I don't think this is the right answer, can someone please help?

Thanks.

 

[Quinzio]

a = g*sin(theta)
= 0.51289 m/s^2

I can't understand how have you imagined that. Can you explain ?

Anyway the problem is quite confusing.
Why do they tell you "making a 3.00 degree angle with the ground" ?

 

[lvslugger36]

Use the range equation for this, since it returns to the same initial height released (theoretically). V^2 Sin (2 theta)/g = Range. We know that the angle of 3 degrees it makes with the ground is the same three degrees it was released at above the horizontal. Hence, we have v^2 (sin 6)/9.8 = 66. Solve for v.

 

[a.d.d92]

@ Quinzio: i have no idea how I did it, I don't even know how to start?

@ lvslugger36: wait how would use the range equation for this? Can't you only use the range equation only if the object lands at the surface it was launched?

 

[lvslugger36]

sorry, if he releases the arrow parallel to the ground it makes half a parabola in its trajectory. Now, draw the other half. You'll see that the x distance of this parabola now is 66 x2. Then, you can follow my last post. If you'd like to stick with this, you'd have to solve for the x component of the speed which is v cos thetha. However, I came up with two alternate solutions. We know that 66 = vt. We know in the y direction dy = 4.9t^2. In terms of the 3 degrees the arrow makes, we know that vy = vx tan 3. So we will use vf^2 = vo^2 + 2ad, in the y-direction. vo = 0. Now, we substitute for the vyfinal with vxtan3. So we get vx^2 (tan3)^2 = 19.6d. We substitute for d, which is 4.9t^2. We then substitute for t, which is 66/vx. Now, we have one unknown, and you can solve. The other is just to use vf = vi + at for the y direction. Vi drops out and we have vxtan 3 = 9.8 (66/vx). You can also solve this way. However, dealing with it as a parabola is easier than the other two ways.

 

[a.d.d92]

Okay I don't really understand why you use "tan" ? and second, Is the range thing in your older post right as well? I'll use that instead

 

[Liquidxlax]

i thought you would need an initial starting height because gravity would take such and such time to make the arrow fall a certain height

 

[lvslugger36]

No, in my older post, i forgot to include the fact that the arrow was released parallel to the ground (ie no initial y velocity). I will re-write the corrected solution right here. If he releases the arrow parallel to the ground it makes half a parabola in its trajectory. Now, draw the other half. You'll see that the x distance of this parabola now is 66 x 2. Now, use the range equation for this, since it returns to the same initial height released. V^2 Sin (2 theta)/g = Range. We know that the angle of 3 degrees the arrow makes with the ground is the same three degrees it was released at above the horizontal. Hence, we have v^2 (sin 6)/9.8 = 66 x 2. Solve for v. Then, find the x-component of that v, which is vcos3.

 

[a.d.d92]

okay I see what you're doing there but (66m x2)? why do multiply it by 2? I'm only drawing half of the parabola:

 

[stevenb]

I like lvslugger's solution, as it is creative. He is basically saying that you can pretend that the arrow is launched upward at a three degree angle, and it will travel twice as far. This allows him to use a well established formula to find the answer.

An alternative method is to use the well known small angle approximation which says that the sine and tangent of a small angle are about equal to that angle in units of radians. Then you can take the derivative of the trajectory and set it equal to the angle in radians. Then solve for the velocity using x=66 m.

 

[lvslugger36]

In order to use the range formula, we would have to draw in the other half of the parabola so that the arrow ends up at the same height as it started (ie a vertical displacement of zero). This means we have the 66 m which is the half parabola it travels, plus the 66 m of the other half parabola that we draw into compete the parabolic trajectory. In other words, we have half a parabola that the arrow travels. The x-distance of this parabola is 66m. Now, we want to complete this parabolic shape by adding in the half parabola the arrow would have taken to get to the point it was released (in other words, from the ground to the starting point of the original half parabola). That x-distance is also 66 m. So in the full parabolic trajectory, the x-displacement is 66 x 2 m.