Did we need to know that?OMG:O
This thread is ver confusing too melol
you can simplify s=ut+0.5a(t^2)
as assuming no are resistance a=0
therefore s=ut
t= time for ball to drop, can be found with same equation
s=ut+0.5a(t^2)
however this time there is no initial speed (u) so
s=0.5a(t^2)
re-arrange
(2s/a)^0.5=t
andThat IS the simple version.
u=100m/s
a=0
m=.002kg (approx)
acc. due to gravity=g= 9.8m/s/s
horizontal distance traveled=s=ut + 1/2a[t^2] => s=100t
just pick a height and apply the same formula, putting the chosen height as s, and using g for the acceleration which gives you the time (t) for it to hit the ground, then stick that into the horizontal formula: that's how far a paintball will travel when fired parallel to the ground. The forces get a little more complex (not much) when you angle it because then you have to split it into vectors and you have an initial speed (u) for the vertical movement with -g as the acceleration. You'd have to do a bit of calculus if you wanted to take air resistance into account because the acceleration would change with velocity (a=-kv)
Thanks.another reasonably easy one to work out is how far will the ball travel before it hits the ground.
assume a hight (measure it) you are holding the gun and assume that you are holding it horizontal....you know the speed the ball is moving horizontally and you can use affect of gravity to work out how long the ball will take to drop to the groud.......then use this time to work out how far the ball will travel in this time. if you want to get a little more involved you can do the same thing but assume the barrel is angled up at 5*.