How is a multi-particle system in classical mechanics solved? A wheel of radius $a$ is rolling along a muddy road with speed $v$.  Particles of mud attached to the wheel are being continuously thrown off from all points of the wheel.  If $v^2> ag$, show that the maximum height above the road attained by the flying mud will be $$h_{\max}=a+\frac{v^2}{2g}+\frac{a^2g}{2v^2}.$$
Explain the origin of the requirement $v^2> ag$.
I'm not even sure where to begin with this kind of problem. If someone could help show how a multiparticle question such as this one is solved that would be greatly appreciated.
 A: You can just set up your equations of motion for a single particle. The initial velocity in the y-directions is given by the y-velocity of a point on the wheel. Once you get that velocity then you solve the problem like a regular projectile motion problem. Recall the tangential speed of a point on a wheel $v = r\omega$ and $\omega$ is the angular speed of the wheel in radians/sec. And use the fact that the tangential velocity must be pointed upwards to be a solution to the problem.
Set up and solve the equation of motion for h, and remember h must be upwards.  Then look at what $v^2 > ag$ implies.
A: Consider a point on the rim of the wheel at an angle $\theta$ above the horizontal. A particle at that point starts at a height $a(1+\sin \theta)$ and has vertical speed $u_v=v \cos \theta$. If it travels ballistically it will therefore reach a maximum height of
$\displaystyle h_{max}=a(1+\sin \theta) + \frac {u_v^2}{2g} = a(1+\sin\theta) +\frac {v^2 \cos^2 \theta}{2g}$
You need to find the value of $\theta$ that maximises $h_{max}$, then substitute this back into the expression for $h_{max}$.
