# Range of an arrow, shot BACKWARDS from a moving car

## 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 car

I 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 you

Edit!!!
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

Last edited:

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gneill
Mentor
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.

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. Im 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?

gneill
Mentor
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...

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 :)

gneill
Mentor
Are the car and arrow velocities in the same direction?

gneill
Mentor
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).

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?

gneill
Mentor
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.

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

gneill
Mentor
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...

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