ok theres two cannons ones at an 90 degree angle and theres another one exactly half of the angle. They are both shot a the same muzzle velocity. Which hit the ground firt. I think its both but some of my friends say the 90 degree will hit first. Any ideas formulas or answers? And if you have the answer could you show some proof?

the half angle hits first, less of its force is pushing up, which means less time before upward velocity=0(when it starts heading back down) which means less time before it hits bottom again.

 

lets say velocity=10f/s.

 

vertical starting velocity on the 90=10f/s

t=time in secs

vertical position =10t-16t^2

when position =0 it is on the ground

 

0=10t-16t^2

0=t(10-16t)

t=0(right when you shoot) or 5/8=.625(when it hits)

 

5/8 sec= impact

 

vertical starting velocity on the 45= 10sin(45)

 

position=10sin(45)t-16t^2

 

0=10sin(45)t-16t^2

0=8.509t-16t^2

0=t(8.509-16t)

t=0 or 8.509/16=.5318 secs

 

90 hits at .625 secs

45 hits at .5318 secs

 

been a few years since physics, im sure the 45 hits first but my exact numbers may be off, anyone got any corrections?

 

edit: doh, -16 t^2 is gravity in feet, 9.8 for meters, there: were going with feet, yay america.