Projectile Motion given angle and distance, find intial velocity

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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 degree angle with the ground.


Homework Equations



x(t)= x(0) + Vx[0]T
V(Y)= Vy[0]Sin(theta) - 1/2gt^2
V(x)= Vx[0]cos(theta)t



The Attempt at a Solution


x(t)= Vx[0]t
58= Vx[0]cos(3)t
t= 58/ Vx[0]cos(3)


I know I should then plug this into the y component equation, but I thrown off since there is no height given. I end up with too many variables. I am very lost on this problem!
 

Answers and Replies

  • #2
Doc Al
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Since the arrow is shot horizontally, the initial vertical speed is zero. Assume that the arrow always points in the direction of its velocity.
 
  • #3
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I am still getting this wrong. I seem to always end up wth 2 unknowns, and no way to plug one into the other equation. This is what I've come up with:

since there is a no initial velocity in the horizontal direction, I got this for the x-component.:
Vx[0]= Vi[x]
=58=Vx[0]

And for the height, which is the other unknown, I got:
H= Vy[0]- .5gt^2

I still do not see a way in which I can solve for the time, and which equation I am trying to use to solve for the velocity. I'm missing something with understanding this problem.
 
  • #4
Doc Al
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since there is a no initial velocity in the horizontal direction, I got this for the x-component.:
Vx[0]= Vi[x]
=58=Vx[0]

58 m is the horizontal distance, not the velocity. Express horizontal distance as a function of time.

And for the height, which is the other unknown, I got:
H= Vy[0]- .5gt^2

Try this: Express vertical distance as a function of time using the average vertical speed.

Combine those two equations with what you know about the angle of the final velocity to solve for Vx.
 
  • #5
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Thank you eveyone for your help, but I am still greatly confused.
These are these equations I have. Can someone tell me where I am making a mistake:
y(t)= y[0] + Vy[0]-.5at^2
then I modified it as follows:
y(t)= Vy[0]sin(theta)-.5at^2

For horizontal distance as a function of time I got:
x(t)= x[0]+ Vx+ .5at^2
I eliminated x initial and the accelartion part and got:
X(t)= Vcos(theta)t
solving for t:
X(t)/Vcos(theta)

I am super confused for this problem. I do not see how I can get a value for t or V because it seems as though I have too many unkowns. Am I supposed to use tangent somewhere? I appreciate any and all help. I just need someone to explain this problem to me in lamest terms.
 
  • #6
Doc Al
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These are these equations I have. Can someone tell me where I am making a mistake:
y(t)= y[0] + Vy[0]-.5at^2
then I modified it as follows:
y(t)= Vy[0]sin(theta)-.5at^2

Let's do something even simpler. Write the final vertical speed Vy as a function of time.

For horizontal distance as a function of time I got:
x(t)= x[0]+ Vx+ .5at^2
I eliminated x initial and the accelartion part and got:
X(t)= Vcos(theta)t
solving for t:
X(t)/Vcos(theta)

Let's make it simpler:
X = Vx t

How are Vx and Vy related?

You'll end up with 3 simple equations that you can noodle around with and solve for Vx.

I am super confused for this problem. I do not see how I can get a value for t or V because it seems as though I have too many unkowns. Am I supposed to use tangent somewhere?
Yes! That will give you the third equation that you need.
 

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