Recent content by putio_d

hahaha that would make it more complicated I guess. Thanks a lot for your help! I really appreciate it

Oh, so I can change it from R = (Vacos(theta) + Vc)*(2Vasin(theta))/g to R = (Vacos(theta) - Vc)*(2Vasin(theta))/g

No. the arrow is moving in the opposite direction. Would that change my R equation?

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...

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...

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...