Mridul B. answered 28d
High School Math & Physics Tutor | Competition & AP Specialist
The first thing to realize is that the initial speed from when the mud leaves the wheel is the same as the speed of the car (v). We rationalize by noting that the speed of the car and any point on the wheel must be the same. And since the instant before the mud was flying, it was on the wheel, it must also have the same speed.
Then we note that we don't know the angle the mud leaves the car. This is a good time to start drawing a diagram. Draw the wheel, setting the origin at the bottom of the wheel. This puts the center of the wheel at (0, a). Then, let's draw a line anywhere from the center to where the mud is flung. We note it must be the top half of the wheel. The top half because, trivially, the mud will go highest when its flung from the top of the wheel. Let's define the angle between our line and the y-axis as Θ. Since we care about height, we isolate the y-velocity of the mud. Doing some angle-chasing, we get vy=v sin(Θ). We also get that the initial position is a+a cos(Θ).
Of the kinematics equations, we use vf2-vi2=2a(xf-xi), as it's the only one without time. We note that at the height, vf=0. We have vi=v sin(Θ), a = g, and xi=a+a cos(Θ). We plug and isolate for xf. Doing so yields
a + a cos(Θ) + (v2sin2(Θ))/(2g). We note that the only variable is Θ. To make things easier, we use sin2(Θ) = 1 - cos2(Θ), as then we only have one trig function. Re-writing, we get: xf=a + a cos(Θ) + v2/(2g) - (v2cos2(Θ))/(2g). We note that this a quadratic in cos(Θ). We use the formula for maxmizing the quadratic and get cos(Θ) = (ag)/v2. This is where the initial condition comes in, as this would not be valid unless v2>ag, like the problem stated. We plug-and-chug, and get the final answer of a max height of: a + v2/(2g) +(ga2)/(2v2).
Sidenote: if we didn't have the initial condition, we would max at Θ = 90. Try proving this to yourself!