Let’s suppose you built a potato cannon and invited some friends to fire it. You became tired of shooting potatos into the Hudson River so you started shooting signs, the occasional New York City rat, and 40-ounce malt liquor glass bottles. You wondered how high the potato could go if you fired it vertically, its maximum range, its muzzle speed, etc. If you enjoy accelerating starchy tubers up to insane speeds keep reading.

Assumption: no air resistance => potato doesn’t have a terminal velocity on its way down => time it takes to go up equals the time it takes to come down.

$$ \begin{align} y &= \textrm{height}\ v{y} &= \textrm{vertical velocity}\ v{0y} &= \textrm{initial vertical velocity} \end{align} $$

gravitational acceleration constant:

$$g = 9.8\frac{m}{s^{2}}$$

Basic motion equations:

$$ v{y} = v{0y} - gt$$

$$y = v_{0y}t - \frac{1}{2}gt^{2}$$

We calculate the initial or muzzle velocity:

$$0=v{0y} - g \cdot 4.75 \Rightarrow v{0y} = 46.6\textrm{m/s}$$

We now plug the initial velocity into the second motion equation above. This lets us calculate the peak height:

$$y = 46.6 \cdot 4.75 - \frac{1}{2} \cdot 9.8 \cdot 4.75^{2} = 110.6 \textrm{m}$$

Knowing the muzzle speed, we can determine the cannon’s maximum range if we angle it at 45 degrees or π/4 radians to the horizon.

$$v{0y} = cos(\frac{\pi}{4}) \cdot v{0} = \frac{\sqrt{2}}{2} \cdot 46.6 = 33\textrm{m/s}$$

$$y = v_{0y}t - \frac{1}{2}gt^{2} = 33t - \frac{1}{2} \cdot 9.8t^{2} = 33t - 4.9t^{2}$$

Solve this quadratic equation for t by setting y equal to 0 gives us $$t = 6.72s$$. Peak height occurs when $$v_{y} = 0$$. We use the vertical motion equation:

$$0 = v_{0y} - gt = 33 -9.8t \Rightarrow t = 3.4s$$

$$y = 33 \cdot 3.36 - \frac{1}{2} \cdot 9.8 \cdot 3.36^{2} = 55.3\textrm{m}$$

The total flight time is simply twice the time for the potato to reach peak height: 6.7s. Finally the horizontal distance is calculated similarly using the initial horizontal velocity and the total flight time:

$$x = v_{0x}t = 33t = 33 \cdot 6.7 = 221.1\textrm{m}$$

Now how do I factor in air resistance and do cool calculations with the pressure inside the combustion chamber?