Where should I go to do it? A football field is definitely too small. The only marked land I can think of would be a driving range and I think of golf as being a high-SES sport, so I wouldn't imagine the owners would let me shoot it there, even if it were after hours.

Thoughts?

The beach?

Try shooting vertically and evaluating how long it takes to reach maximum height.

a=-g
v=v0-gt0

max height: v=0

v0=gt0

So take your time, multiply it by 9.8 and you have the initial speed.

We will neglect the fluid resistance and stuff to make it easier and with air all around, you will get a more or less good approximation.

Now when you shoot it with an Ô angle from the ground, the trajectory will be:

a=-g
v(t)=v0cos(Ô)-gt
x(t)=x0+v0cos(Ô)t-(1/2)gt²

x0=0 from your point of view so:

x(t)=v0cos(Ô)t-(1/2)gt²

now we have to find t(max) so that x(t)=0

Delta=v0²cos(Ô)²
s=(-b+sqrt(delta))/2a
s=-2b (since delta is b²)/2a
s=-b/a

t(max)=v0cos(Ô)/(1/2)g
t(max)=2t0cos(Ô)

so after t(max)=2t0cos(Ô), the potato touches the ground again

On the horizontal axis, the potato moves v0cos(Ô)t, so if we replace:

(xi is on the horizontal axis)
xi(max)=v0cos(Ô)2t0cos(Ô)
xi(max)=2t0v0cos(Ô)²=2t0gt0cos(Ô)²

xi(max)=19.6t0²cos(Ô)²

The distance at which the potato lands from you is :

xi(max)=19.6t0²cos(Ô)²

with t0 the time it takes to reach its high point when shot vertically and Ô the angle at which you shoot the potato with from the ground

Have fun !! =D

@ RubixFreakGreg: Clever stuff, I like the way you think.

Mind you, it would probably be a lot more fun just to smash it into things, see how much damage can be done...

well if we're talking smashing you can always calculate drag, but it's out of my league to post something correct, so in short range you get v=v0, which could go from 50 to 900 (world record, 3 times faster than sound) m/s which is basically, for a 200g potato: 1/2*0.2*v0²=0.1v0² J of kinetic energy.

So for 50-900 m/s, you get Ec (kinetic energy, but i'm french so sorry for the different letters ^^) varying from Ec(min)=5 J to Ec(max)=90 J

The strength of your potato launcher is equivalent to the strenght of ________ at __ to __ [unit of speed]

Queen Mary 2: 0,000257-0.00109 m/s
Electron: 4.68*10^10-3.31*10^15 times the speed of light
Car: 0.1-0.4243 m/s

I like those stats ^^

I just got a pedometer. It goes an avg of about 400ft