If I shoot a ball out of a cannon with muzzle velocity of 100 fps at 45 degree angle, how do I compute how far I shoot it? Ignore air resistance for this.
and thus x=t*v^2/g instead of v^2/g
Very well :)Sure t=2Vy/a - thus 2x70.7/32=4.41875 secs
which then gives 312ft for horizontal distance
from 4.41875x70.7. This all from a muzzle velocity of 100 fps
aimed at a 42degree angle
is that correct?
Your answer is correctif horizontal flight time is t=2xVxcos(45)/a
distance horizontally is txVxcos(45)
combining them is 2xv^2*cos^2(45)/a and for
100fps initial velocity on level ground, this means a distance of 312.5 feet
Note that (from above) v^2sin(2x45)/g also works and gives 312,5 ft
If this correct, thanks to all for all your help.
if this is wrong, where am I wrong?