Projectile Motion Question involving a ball and a ramp inclined at an angle The question is to finde the initial horizontal velocity of the ball at end of the ramp, where it is released.
I know how to do this using gravitational potential energy and kinetic energy ($v=\sqrt{2gh}$), assuming all potential energy is converted into kinetic energy but the question is asking me to find the error in the experiment.
It gives me the values on $y$ as the ramp is lifted up (it is lifted up about 10 cm each time) and the corresponding values of $x^2$. I drew a graph of $x^2$ vs $y$ and found the gradient.
How can I use this to find the experimental initial horizontal velocity?
This sketch shows what the experiment looks like: 

 A: 
Disclaimer: Since this is a homework style problem, I'm not going to write all the details.
Writing down the projectile's equation, and replacing $\theta$ we arrive at the equation:
$$-\frac{x^2}{\frac{4 g (h-y) \left(l^2-(y-h)^2\right)}{l^2}}-\frac{x (y-h)}{\sqrt{l^2-(y-h)^2}}+h=0$$
Solving this for $x$, and simplifying the huge expressions(perhaps by something like Mathematica); we find:
$$x\to \frac{ \left( 2 \sqrt{g (h-y) (h+l-y)^2 (-h+l+y)^2 \left(g (h-y)^3+h l^2\right)}-2 g (h-y)^2 (h-l-y) (h+l-y)\right)}{l^2 \sqrt{-(h-l-y) (h+l-y)}}$$
Squaring:
$$\Rightarrow x^2=\frac{1}{l^4}4 g (y-h) \left(-2 h \sqrt{g (h-y) (h+l-y)^2 (-h+l+y)^2 \left(g (h-y)^3+h l^2\right)}+2 y \sqrt{g (h-y) (h+l-y)^2 (-h+l+y)^2 \left(g (h-y)^3+h l^2\right)}+2 g (h-y)^3 (h-l-y) (h+l-y)+h l^2 (h-l-y) (h+l-y)\right)$$
In general if we have a function of several independent parameters(i.e. $f(x_1, x_2, \cdots , x_n)$), and we want to calculate the error in $f$, we would write:
$$\delta f = \sqrt{\sum_i \left( \frac{\partial f}{\partial x_i}\delta x_i \right)^2}$$
In this case $x^2$ only depens on $y$, and its error comes from the error in $y$:
$$\delta (x^2) = \left| \frac{\partial (x^2)}{\partial y} \delta y\right| = x^2 \left|\frac{ \left(-g (h-y)^3 \left(3 (h-y)^2-l^2\right)+3 h \sqrt{g (h-y) (h+l-y)^2 (-h+l+y)^2 \left(g (h-y)^3+h l^2\right)}-3 y \sqrt{g (h-y) (h+l-y)^2 (-h+l+y)^2 \left(g (h-y)^3+h l^2\right)}+h l^2 \left(l^2-3 (h-y)^2\right) \right)}{(h-y) (h-l-y) (h+l-y) \left(g (h-y)^3+h l^2\right)}  \delta y\right|$$
It doesn't even fit properly! Anyway, putting back all the corresponding numbers, we find:
$$\frac{\delta (x^2)}{x^2} \approx 0.52 \delta y$$
Which looks like a decent result to me. Note, this final equation is only true for values given in this link, namely $h=9.7 \text{cm}, y=17.5 \text{cm}, l=60\text{cm} \ \text{and}\  g=980 \text{cm}\text{.s}^{-2}$.
A: Vertically:
$\ddot{y}=a$
$\frac{d\dot{y}}{dt}=a\therefore \int d\dot{y}=\int adt\therefore \dot{y}=u\sin\theta +at$
$\frac{dy}{dt}=u\sin\theta +at\therefore \int dy=\int \left ( u\sin\theta +at \right )dt\therefore y=u\sin\theta t+\frac{a}{2}t^{2}$
Horizontally:
$\ddot{x}=0$
$\frac{d\dot{x}}{dt}=0\therefore \int d\dot{x}=\int 0dt\therefore \dot{x}=u\cos\theta$
$\frac{dx}{dt}=u\cos\theta\therefore x=u\cos\theta t$
We know the gradient of the graph of $\frac{x^{2}}{y}=m$ ($m$ is the gradient, measured in $cm$)
Sub $x=u\cos\theta t$ and $y=\frac{a}{2}t^{2}$
$\therefore u\cos\theta=\sqrt{\frac{\frac{a}{2}m}{100}}m/s$
Theoretically, $u\cos\theta$ can be found by considering the conversion of gravitational potential energy to kinetic energy, assuming no energy loses
$\therefore mah=\frac{1}{2}m\left (u\cos\theta   \right )^{2}\therefore u\cos\theta=\sqrt{2ah}$
A: There is an 'error' in your reasoning. The velocity of the ball at the end of the ramp is not given by $v=\sqrt{2gh}$. This assumes that the ball slides down the ramp. Presumably it rolls and gains rotational as well as translational kinetic energy. If the ball has uniform density and rolls without slipping then the velocity after rolling down an incline is $v=\sqrt{\frac{10}{7}gh}$, which is smaller. If the ramp is quite steep there could be a mixture of rolling and sliding, so the launch velocity will be somewhere in between the two values. 
