Difficult Projectile Motion Arrow Problem

I understand it much better now. In summary, the original question asked for the speed of an arrow shot from a bow when it stuck in the ground 59.0 m away at a 3.00 degree angle with the ground. By using the kinematic equations and position, velocity, and acceleration definitions, the horizontal speed of the arrow was found to be approximately 74.68 m/s. This was found by solving for vx using the equation vy/vx = gx/vx^2 and substituting the values of vy and x.
  • #1
JoshMP
38
0

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

Homework Equations



Kinematic equations
Position, velocity, and acceleration definitions

The Attempt at a Solution



I've spent an hour on this problem. I know it's long and windy but any help would be great!

Here has been my work so far; perhaps someone can show me where I've gone wrong (?) or if I'm even on the right track.

First, I used the 3 degree angle and said that dy/dx= tan (3 degrees) when the arrow hits the ground, so dy/dx=.0524 (at time t, when the arrow hits the ground).

Then I said that dx/dt is constant, since there is no horizontal acceleration. The integral of dx/dt is position, so the integral of dx/dt = 59 m.

I then multiplied dy/dx by dx/dt to give me dy/dt. Since dy/dx=.0524 and dx/dt= vx, I said that dy/dt=.0524(vx).

With that expression, I integrated both sides. The integral of dy/dt is dy, change in position of y. The integral of .0524(dx/dt) is .0524(dx), which is equal to .0524(59), or 3.0916.

In other words, if my mathematics has been correct up to this point, THE CHANGE IN POSITION IN THE Y DIMENSION IS 3.0916.

I then plugged 3.0916 into s= .5(a)t^2 and solved for t. I got .79 seconds for t. Using this, I used the same equation for change in x position. Written out, this looks like 59m= (vix)(.79s). I got vix=74.68 m/s. The answer is wrong.

Can someone please help me? I have to know how to do this kind of problem.

Thanks!
 
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  • #2
JoshMP said:
I then multiplied dy/dx by dx/dt to give me dy/dt. Since dy/dx=.0524 and dx/dt= vx, I said that dy/dt=.0524(vx).

dy/dx is only equal to 0.0524 at the very end of the arrow's journey. Since you're trying to apply dy/dt to the entire trajectory, you can't use dy/dx=0.0524.

You've found the arrow's horizontal speed. That's your answer; the arrow has no horizontal acceleration, so that speed can't possibly change from when it was fired.
 
  • #3
ideasrule said:
dy/dx is only equal to 0.0524 at the very end of the arrow's journey. Since you're trying to apply dy/dt to the entire trajectory, you can't use dy/dx=0.0524.

You've found the arrow's horizontal speed. That's your answer; the arrow has no horizontal acceleration, so that speed can't possibly change from when it was fired.

I found the horizontal speed using dy/dt=.0524. If I can't use that expression for the entire trajectory, how do I find the horizontal speed?
 
  • #4
Is there a way I could use the information of vertical acceleration? Like, d^2y/dt= vertical acceleration= -9.8 m/s^2?
 
  • #5
I integrated both sides. The integral of dy/dt is dy, change in position of y. The integral of .0524(dx/dt) is .0524(dx), which is equal to .0524(59), or 3.0916
This step is not correct. You have to write it as
dy = 0.0524*vx*dt. Integrate it . You get
y = 0.0524*vx*t + C. When t = 0 , y = yo. The better method will be
-y = -1/2*g*t^2. Because initial vy = o
t = x/vx.
So -y = - 1/2*g*x^2/vx^2
vy/vx = gx/vx^2. Substitute the value of vy. You get
0.0524*vx^2 = g*x. Now solve for vx.
 
Last edited:
  • #6
rl.bhat said:
vy =dy/dx = gx/vx^2.

You lost me right there. How does Vy= dy/dx? Shouldn't Vy=dy/dt?
 
  • #7
JoshMP said:
You lost me right there. How does Vy= dy/dx? Shouldn't Vy=dy/dt?
Ok.
Then dy/dx= dy/dt*dt/dx = gx/vx^2
vy/vx = gx/vx^2
0.0524*vx^2 = gx. Find vx.
 
  • #8
=) Thank you so much!
 

1. What is projectile motion?

Projectile motion is the motion of an object through the air, such as a ball or an arrow, that is subject to the force of gravity. It follows a curved path known as a parabola.

2. What makes the "difficult projectile motion arrow problem" challenging?

The "difficult projectile motion arrow problem" refers to a specific scenario where an arrow is fired at an angle from a certain height and distance, and the goal is to determine the initial velocity and angle needed to hit a specific target. This problem is difficult because it requires applying various equations and concepts from physics, such as kinematics and vectors, to solve for multiple unknown variables.

3. How can I solve the "difficult projectile motion arrow problem"?

To solve the "difficult projectile motion arrow problem", you will need to use equations for projectile motion, such as the range equation and the maximum height equation, along with trigonometry to analyze the initial velocity and angle needed for the arrow to hit the target. It is also helpful to break down the problem into smaller parts and use a step-by-step approach.

4. What are some common mistakes when solving the "difficult projectile motion arrow problem"?

One common mistake is forgetting to take into account the initial height and distance of the arrow from the target. Another mistake is not properly setting up the equations or using the wrong formulas. It is also important to pay attention to units and make sure they are consistent throughout the problem.

5. How is projectile motion used in real life?

Projectile motion is used in various real-life scenarios, such as sports like basketball and soccer, where players need to accurately shoot or kick a ball into a goal. It is also used in military and defense applications, such as calculating the trajectory of missiles and artillery. Additionally, understanding projectile motion is important in engineering and designing structures or vehicles that need to withstand the force of gravity.

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