- #1

- 7

- 0

## 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: