Projectile motion of arrow shot parallel to ground

[cmkluza]

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 58.0 m away, making a 3.00° angle with the ground. How fast was the arrow shot?

Homework Equations

Kinematic equations are useful, I think.
$d = v_it + \frac{1}{2}at^2$

The Attempt at a Solution

As I wrote this out I realized a huge flaw in my logic - I assumed there to be no initial vertical velocity because the arrow was shot parallel, but then assumed the path the arrow took was a triangle right from the start (which would imply it's being shot at an angle, not parallel to the ground). Thus, my work below is definitely incorrect, but it's all I've got so far so I'll leave it. Where can I start with the given information?

I started by forming a triangle (angles 90, 87, and 3 degrees). Using the sine rule, I found the vertical height the arrow was shot from:
$\frac{h}{\sin(3)} = \frac{58}{\sin(87)} \longrightarrow h = \frac{58\sin(3)}{\sin(87)} \approx 3.04$m
It stated the arrow was shot parallel to the ground, so I assume an initial vertical velocity of zero. I then use acceleration of gravity (9.81) to try to solve for total time the arrow is in the air:
$d = v_it + \frac{1}{2}at^2 \longrightarrow 3.04 = \frac{1}{2}(9.81)t^2 \longrightarrow t^2 = \frac{2 \times 3.04}{9.81} \approx 0.62 \longrightarrow t = \sqrt{0.62} \approx 0.79$
With this, I assume I can solve for initial horizontal velocity, which (as it was shot parallel to the ground) should be the entire initial velocity, and there should be no horizontal acceleration:
$d = v_it + \frac{1}{2}at^2 \longrightarrow 58 = v_i(0.79) \longrightarrow v_i = \frac{58}{0.79} \approx 73.68$.
However, this is not the answer. Any hints on what I'm doing wrong?

 

[berkeman]

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 58.0 m away, making a 3.00° angle with the ground. How fast was the arrow shot?

Homework Equations

Kinematic equations are useful, I think.
$d = v_it + \frac{1}{2}at^2$

The Attempt at a Solution

As I wrote this out I realized a huge flaw in my logic - I assumed there to be no initial vertical velocity because the arrow was shot parallel, but then assumed the path the arrow took was a triangle right from the start (which would imply it's being shot at an angle, not parallel to the ground). Thus, my work below is definitely incorrect, but it's all I've got so far so I'll leave it. Where can I start with the given information?

I started by forming a triangle (angles 90, 87, and 3 degrees). Using the sine rule, I found the vertical height the arrow was shot from:
$\frac{h}{\sin(3)} = \frac{58}{\sin(87)} \longrightarrow h = \frac{58\sin(3)}{\sin(87)} \approx 3.04$m
It stated the arrow was shot parallel to the ground, so I assume an initial vertical velocity of zero. I then use acceleration of gravity (9.81) to try to solve for total time the arrow is in the air:
$d = v_it + \frac{1}{2}at^2 \longrightarrow 3.04 = \frac{1}{2}(9.81)t^2 \longrightarrow t^2 = \frac{2 \times 3.04}{9.81} \approx 0.62 \longrightarrow t = \sqrt{0.62} \approx 0.79$
With this, I assume I can solve for initial horizontal velocity, which (as it was shot parallel to the ground) should be the entire initial velocity, and there should be no horizontal acceleration:
$d = v_it + \frac{1}{2}at^2 \longrightarrow 58 = v_i(0.79) \longrightarrow v_i = \frac{58}{0.79} \approx 73.68$.
However, this is not the answer. Any hints on what I'm doing wrong?

The arrow follows a parabolic path. Were you given the initial height off the ground?

 

[cmkluza]

The arrow follows a parabolic path. Were you given the initial height off the ground?

That makes sense, but I didn't know if I was supposed to make some assumptions, since I didn't know what I could do with the given information.

I copied and pasted all the information that problem statement had, so all the information we're given is initial shot is parallel to the ground, travels 58 m, and the arrow makes a 3 degree angle with ground.

 

[berkeman]

That makes sense, but I didn't know if I was supposed to make some assumptions, since I didn't know what I could do with the given information.

I copied and pasted all the information that problem statement had, so all the information we're given is initial shot is parallel to the ground, travels 58 m, and the arrow makes a 3 degree angle with ground.

Well that's a bit harder, but still workable, I think. You need to calculate the parabolic path that intersects with the ground at that angle that far away.

 

[cmkluza]

Well that's a bit harder, but still workable, I think. You need to calculate the parabolic path that intersects with the ground at that angle that far away.

I'll be honest, I have no idea where to start in order to calculate that, I've never had to work with parabolas in that way before.

 

[kuruman]

The triangle method would work if the arrow traveled in a straight line. Your assumption that the arrow was initially shot in the horizontal direction is correct. By the time the arrow hits the ground it has acquired a vertical component as well. The right triangle you should be thinking about is a final velocity triangle. That's the one that has the 3^o angle as part of it.

 

[berkeman]

I'll be honest, I have no idea where to start in order to calculate that, I've never had to work with parabolas in that way before.

What derivative would you take to get the angle of the parabola with the ground at any point x away from the peak (the archer is at the origin)...

 

[gneill]

This question is all about employing some of the details of projectile motion and how they're related. It's a bit tricky to recognize where to start, but if you consider the scant given information clearly the impact angle with the ground is important. What do you know about the impact angle (besides its value that is)?

 

[berkeman]

The right triangle you should be thinking about is a final velocity triangle.

What derivative would you take to get the angle

kuruman's hint is better -- you probably don't need to use a derivative to answer this question.

 

[cmkluza]

The triangle method would work if the arrow traveled in a straight line. Your assumption that the arrow was initially shot in the horizontal direction is correct. By the time the arrow hits the ground it has acquired a vertical component as well. The right triangle you should be thinking about is a final velocity triangle. That's the one that has the 3^o angle as part of it.

I'm still not sure what I'm doing wrong. I managed to work out the following, but it's still incorrect:
In the horizontal direction:
$a=0$
$v_{ix}=v_{fx}=v_f\sin(3)$
$d=58$
I plugged these into the following equation:
$d=v_it+\frac{1}{2}at^2 \longrightarrow 58=v_f\sin(3)t \longrightarrow v_f=\frac{58}{\sin(3)t}$
In the vertical direction:
$a=9.81$
$v_{iy}=0$
$v_{fy}=v_f\cos(3)$
I used the above value for $v_f$ in the following equation to find $t$:
$v_{fy}=v_{iy}+at \longrightarrow v_f\cos(3)=at \longrightarrow (\frac{58}{\sin(3)t})\cos(3)=9.81t \longrightarrow \frac{58\cos(3)}{9.81\sin(3)}=t^2 \longrightarrow t \approx 10.6$
With a value for time, and since I don't need to worry about a vertical initial velocity, I can just use values from the horizontal direction with this time to get an initial velocity:
$d=v_{ix}t+\frac{1}{2}at^2 \longrightarrow 58 = v_{ix}(10.6) \longrightarrow v_{ix}=v_i \approx 5.46$.

And yet this is still incorrect. What am I doing wrong now? Have I mixed up some vertical and horizontal values where I shouldn't have?

 

[berkeman]

vix=vfx=vfsin(3)vix=vfx=vfsin⁡(3)v_{ix}=v_{fx}=v_f\sin(3)

That looks wrong. The sin() function is almost zero for small angles, but the arrow's horizontal velocity should stay unchanged through the light.

The right triangle you should be thinking about is a final velocity triangle.

Use kuruman's hint, and draw the right triangle for the final vertical and horizontal velocities as the arrow strikes the ground. Then write the equation for the vertical velocity as a function of time. Then can you see how you can solve this?

 

[kuruman]

Use kuruman's hint, and draw the right triangle for the final vertical and horizontal velocities as the arrow strikes the ground. Then write the equation for the vertical velocity as a function of time. Then can you see how you can solve this?

Just about what I was going to say. Remember, in this problem, the horizontal component of the velocity is the initial speed and stays constant until the arrow hits the ground.

 

[cmkluza]

That looks wrong. The sin() function is almost zero for small angles, but the arrow's horizontal velocity should stay unchanged through the light.

Alright, I'm terrible at getting sines and cosines correct. I had drawn it out and originally used cosine for horizontal movement and sine for vertical, but one of my results was off, and for some reason I thought I needed to switch these. I understand now.

Just about what I was going to say. Remember, in this problem, the horizontal component of the velocity is the initial speed and stays constant until the arrow hits the ground.

Not sure how I mixed it up, but I understand what you're saying now. Essentially, I can go back to my equation solving for time in my previous post and properly substitute in sines and cosines to get the following:
$t = \sqrt{\frac{58\sin(3)}{9.81\cos(3)}} \approx 0.56$
Then I can substitute the proper value of time to get the proper value for initial velocity:
$58 = v_i(0.56) \longrightarrow v_i = \frac{58}{0.56} \approx 104.2$

Thanks for all the help!

 

[berkeman]

Alright, I'm terrible at getting sines and cosines correct. I had drawn it out and originally used cosine for horizontal movement and sine for vertical, but one of my results was off, and for some reason I thought I needed to switch these. I understand now.

Not sure how I mixed it up, but I understand what you're saying now. Essentially, I can go back to my equation solving for time in my previous post and properly substitute in sines and cosines to get the following:
$t = \sqrt{\frac{58\sin(3)}{9.81\cos(3)}} \approx 0.56$
Then I can substitute the proper value of time to get the proper value for initial velocity:
$58 = v_i(0.56) \longrightarrow v_i = \frac{58}{0.56} \approx 104.2$

Thanks for all the help!

Hmm, I get a different answer. Do you know if your answer is correct? Can you show your steps again to get to that equation?

See next post...

 

[berkeman]

Nevermind, I found my error. I believe your solution is correct.