Range of an arrow, shot BACKWARDS from a moving car

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In summary: Whenever two velocities are added, you have to use vector addition. So, in this case you have to use vector subtraction. The x-component of the velocity of the arrow with respect to the ground will be Va*cos(θ). The x-component of the velocity of the car with respect to the ground will be Vc. Add those two vectors to get the x-component of the velocity of the arrow with respect to the car. Then, add that to Vc to get the x-component of the arrow's velocity with respect to the ground. That will be the quantity you use in your x-displacement equation. You'll have to handle the y-component of the arrow's velocity separately.
  • #1
putio_d
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Homework Statement


Disregard aerodynamic effects and the height of the person and car.

If the speed of the arrow is Va and that of the car is Vc, and the angle above the horizontal at which the person aims is theta, what is the range, when it is measured On The Ground from where it was shot to where it lands.

Then, if Va = 50m/s , Vc= 18m/s , g=9.81m/ss, what should the angle be to achieve the maximum range?

Finally, what is the maximum ground to ground range, of the same bow, standing on the ground? Meaning, how much of the range is lost when it is shot from a moving car, instead of from rest?

Homework Equations



R = Vo^2(sin2theta)/g

The Attempt at a Solution


So, I am guessing the new equation is R=(Va-Vc)^2sin2theta/g

Also, the angle would have to be 45degrees for it to achieve maximum range.

When I use the given from above, and plug in the modified formula I came up with, I get that the maximum range is 104 m.

And if from rest, I get R = 255 m
Therefore, 151 meters were given up, by shooting the arrow off a moving carI feel like this solution is way too simple and I have missed something. I know that the maximum range is achieved when the angle is 45degrees but I am not sure if it is the same when it is shot backwards off of a moving car. And if it is not, then all of my calculations are wrong.

Any input is much appreciated! Thank youEdit!
Upon looking at it again, i could split the velocities into horizontal and vertical
Therefore, if R= (2/g) * {[Vasin(theta)]*[(Vacostheta - Vc)]}
and when we plug in the given, and assume theta is 45degrees for maximum range
I get R = 125m

Then, from rest it will remain the same R = 255m

So, we have lost 130m

Which one is correct? If any :D
 
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  • #2
The angle required for the maximum range for the moving bow will not be 45 degrees; it'll be less: closer to 35 degrees (I'll maintain the suspense and not give away the precise value!).

I think you'll have to calculate that angle and the resultant range for the moving bow.
 
  • #3
gneill said:
The angle required for the maximum range for the moving bow will not be 45 degrees; it'll be less: closer to 35 degrees (I'll maintain the suspense and not give away the precise value!).

I think you'll have to calculate that angle and the resultant range for the moving bow.

So, would that mean that my equation for R is incorrect? Could you possibly tell me how I can fix it. I am not sure what the bow has to do with it, since we are finding the range from the point of where it was shot to where it lands, on the ground.

Also, I appreciate the suspense, but how do I go about even calculating the angle to end up with about 35 degrees?
 
  • #4
Your equation for R is incorrect, because the angle affects the range differently for the moving bow.

You'll have to go back to the usual equations of motion and work out the range for the moving bow.

Assuming that the arrow velocity is with respect to the bow (it wasn't explicitly stated), then you can write the Vx and Vy components for the velocity of the arrow with respect to the ground for the instant that the arrow is launched.

Find the time of flight from the independent equation of motion for the y-direction. From that you can write an equation for the range (x displacement) and then maximize that w.r.t. the angle.

It's perhaps not as bad as it sounds...
 
  • #5
gneill said:
Your equation for R is incorrect, because the angle affects the range differently for the moving bow.

You'll have to go back to the usual equations of motion and work out the range for the moving bow.

Assuming that the arrow velocity is with respect to the bow (it wasn't explicitly stated), then you can write the Vx and Vy components for the velocity of the arrow with respect to the ground for the instant that the arrow is launched.

Find the time of flight from the independent equation of motion for the y-direction. From that you can write an equation for the range (x displacement) and then maximize that w.r.t. the angle.

It's perhaps not as bad as it sounds...

I edited my original post and have done exactly that, perhaps you did not see that.

So, to kind of write out my train of thought, since i did not actually write it out originally
If Time of flight = 2Va * sin(theta)/g

And x = (Vacos(theta) + Vc)t
when x = R and t=time of flight

then, R = (Vacos(theta) + Vc)*(2Vasin(theta))/g, which is what I put when I edited it originally.

Is R wrong again and where is my mistake?

However, i have no idea how I can maximize that wrt to the angle. Could you possibly guide me through it? Thanks :)
 
  • #6
Are the car and arrow velocities in the same direction?
 
  • #7
To maximize the range, given your expression, first differentiate with respect to θ and set equal to zero. Your resulting equation should have several cos(θ) terms and a sin2(θ) term. replace that last term with the usual 1 - cos2(θ) substitution. Solve for cos(θ).
 
  • #8
gneill said:
Are the car and arrow velocities in the same direction?
No. the arrow is moving in the opposite direction. Would that change my R equation?
 
  • #9
putio_d said:
No. the arrow is moving in the opposite direction. Would that change my R equation?

Well, the net x-component of the velocity should reflect the sum of the arrow's component and the car's component...

One of them should be negative.
 
  • #10
gneill said:
Well, the net x-component of the velocity should reflect the sum of the arrow's component and the car's component...

One of them should be negative.

Oh, so I can change it from
R = (Vacos(theta) + Vc)*(2Vasin(theta))/g
to
R = (Vacos(theta) - Vc)*(2Vasin(theta))/g
 
  • #11
putio_d said:
Oh, so I can change it from
R = (Vacos(theta) + Vc)*(2Vasin(theta))/g
to
R = (Vacos(theta) - Vc)*(2Vasin(theta))/g

That'll do. It's just a matter of keeping track of the various directions that things are moving and how they effect the thing being calculated. I suppose one could fire the arrow "forward" with an angle greater than 90 degrees...
 
  • #12
gneill said:
That'll do. It's just a matter of keeping track of the various directions that things are moving and how they effect the thing being calculated. I suppose one could fire the arrow "forward" with an angle greater than 90 degrees...

hahaha that would make it more complicated I guess.

Thanks a lot for your help! I really appreciate it
 

What is the range of an arrow shot backwards from a moving car?

The range of an arrow shot backwards from a moving car depends on several factors such as the speed of the car, the angle at which the arrow is shot, the weight and design of the arrow, and air resistance. However, on average, the range can be anywhere between 100-200 feet.

How does the speed of the car affect the range of an arrow shot backwards?

The speed of the car plays a crucial role in determining the range of an arrow shot backwards. The faster the car is moving, the farther the arrow will travel. This is because the car's speed adds to the initial velocity of the arrow, giving it more distance to cover.

Does the angle at which the arrow is shot affect its range?

Yes, the angle at which the arrow is shot does affect its range. If the arrow is shot at a higher angle, it will travel a shorter distance. On the other hand, if the arrow is shot at a lower angle, it will cover a longer distance. This is because the higher angle reduces the horizontal component of the arrow's velocity, while the lower angle increases it.

What role does air resistance play in the range of an arrow shot backwards from a moving car?

Air resistance is a significant factor that affects the range of an arrow shot backwards from a moving car. As the arrow travels through the air, it experiences drag, which slows it down and reduces its range. This is why lighter and more aerodynamic arrows tend to travel farther.

Can the range of an arrow shot backwards from a moving car be increased?

Yes, the range of an arrow shot backwards from a moving car can be increased by adjusting the factors that affect it. For example, increasing the speed of the car, using a lighter and more aerodynamic arrow, and shooting at a lower angle can all help increase the range. However, it is essential to keep safety precautions in mind when attempting to increase the range of an arrow shot from a moving vehicle.

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