I have a question about dimensions analysis of a horizontal throw:
A ball is thrown from origin where the $x$-axis points in the horizontal direction and $y$-axis vertically upwards. The ball is thrown with a velocity $v_0$ at the time $t=0$. The starting angle of the throw can be described with $\theta$.
We have two equations: $$x(t) = v_0 t \cos\theta$$ and $$y(t) = v_0 t \sin\theta−\frac12 gt^2$$
In the first part of the task you are asked to find $t_m$ such that $y(t_m)$ is equal to $0$ and then find $x(t_m)$
After what I have calculated:
$$(t_m) = \frac{2v_0\sin\theta}{g} $$
and
$$x_m = x(t_m) = \frac{2v_0^2\sin\theta\cos\theta}{g}$$
You are now asked to introduce 3 dimensionless variables $x^∗$, $y^∗$ and $t^∗$ for x, y and t and scale them with respect to $x_m$ for length and $t_m$ for time.
Do anyone have some tips regarding how I should precede this task?
What the problem asks is to write $$x^*=\frac x{x_m}\\t^*=\frac t{t_m}\\y^*=\frac y{x_m}$$ Then you need to rewrite your original equations in terms of $x^*, y^*, t^*$. For example, your first equation becomes $$x^*x_m=v_0t^*t_m\cos\theta$$ By plugging in the values for your $x_m$ and $t_m$, you get $$x^*\frac{2v_0^2\sin\theta\cos\theta}g=v_0t^*\frac{2v_0\sin\theta}g\cos\theta$$ This simplifies to $$x^*=t^*$$This really means that the fraction of time is equal to the fraction of the horizontal distance.
I will let you figure the other equation, which should relate $y^*$ and $t^*$. It will be a quadratic with roots at $t^*=0$ and $t^*=1$.