# Difficult Projectile Motion Arrow Problem

1. Sep 10, 2009

### JoshMP

1. The problem statement, all variables and given/known data
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.

2. Relevant equations

Kinematic equations
Position, velocity, and acceleration definitions

3. 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.

Thanks!

2. Sep 10, 2009

### ideasrule

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. Sep 10, 2009

### JoshMP

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. Sep 10, 2009

### JoshMP

Is there a way I could use the information of vertical acceleration? Like, d^2y/dt= vertical acceleration= -9.8 m/s^2?

5. Sep 10, 2009

### rl.bhat

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: Sep 10, 2009
6. Sep 10, 2009

### JoshMP

You lost me right there. How does Vy= dy/dx? Shouldn't Vy=dy/dt?

7. Sep 10, 2009

### rl.bhat

Ok.
Then dy/dx= dy/dt*dt/dx = gx/vx^2
vy/vx = gx/vx^2
0.0524*vx^2 = gx. Find vx.

8. Sep 10, 2009

### JoshMP

=) Thank you so much!