$$ \definecolor{answer}{RGB}{0,150,0} $$

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NXT Cannon Velocity

In the video above, I show how to calculate the cannon velocity. In my experiment, I used receipt Impact paper so that when the ball lands on the top carbon paper, a mark is made on the paper beneath it. I could have used a carbon paper sheet, but I would have to supply the paper beneath it.
In this experiment the cannon velocity was 4.3 m/s. If I remove the barrel (used to make the ball travel more accurately), the velocity would be greater.

The way you determine the cannon velocity is dependant on the time it takes to drop the projectile and the distance the projectile is able to shoot. First you shoot the cannon at a horizontally zero degree angle. Then you measure the distance from the base of the cannon to where the projectile first made impact with the surface. That distance can be called \(x\). Next you need to calculate the time. A projectile motion formula is

$$ y - y_o = v_o t + {g t^2 \over 2} \tag{1}$$

If you consider that \(y-y_o\)is your height \(y\) and if you consider that \(v_o\) is zero since we are shooting horizontally, you can say that

$$ y = 0 + \frac{gt^2 }{2} $$

\(y-y_o\) and \(y\) = height, \(v_o\) = zero, \(g\) = gravity, \(t\) = time, \(v\) = velocity. When you solve for time \(t\), you'll have the equation of

$$t = \sqrt {2y \over g} \tag{2}$$

If we consider the projectile formula for horizontal velocity, we have:

$$ x - x_o = v_o t + {a t^2 \over 2} \tag{3} $$

If we consider \(x - x_0\) is distance \(x\) and since we have no horizontal acceleration or deceleration, the above formula simplifies to:

$$ x=v_o t \\ \frac{x}{t} = \frac{v_o t}{t} \\ v_o = \frac {x}{t} \tag{4} $$

Substituting time \(t\) from equation (2) into equation (4) we end up with the following velocity equation

$$ \begin{align*} v_o &= \frac {x}{\sqrt {2y \over g}} = \frac{\frac {x}{1} } {\frac{\sqrt {2y}}{\sqrt{g}}} = \frac {x}{1} {\frac {\sqrt{g}} {\sqrt {2y}}} \\ &= \sqrt {g x^2 \over 2 y} \end{align*} $$

The final answer is shown below.

$$ \textcolor{answer}{ v = \sqrt {g x^2 \over 2 y} } $$