We are asked to consider a system in which a brick is being thrown from the ground level at some angle $\theta$. When the brick hits the ground afterr travelling through air, it slides
on the ground, which has a coefficient of friction of $\mu$. If we assume that as the brick travels through the air, the bottom-facing side of the brick remains parallel with the ground, and that the brick
collides completely inelastically with the ground, we must
find the angle at which we should throw the block to maximize the total horizontal distance it travels before coming to rest.
Morin gives us a small (but useful) hint at the end of the problem statement: think in terms of impulse. We will do exactly that! Consider what happens when the brick collides with the floor: for a given amount of time, a force
is appliedto the brick such that its vertical velocity is reduced to $0$. If we throw a brick with some initial velocity $v_0$, at angle $\theta$, then just before it hits the ground, the angle that the velocity
vector makes with the horizontal will also be $\theta$. Thus, we can deduce that the block's vertical momentum before the collision is given by $m v_0 \sin \theta$, and after the collision, it is $0$. Thus, we know that
an impulse is applied to the block. The net force acting in the vertical direction on this block is given by:
$$F_{\text{Net}} \ = \ N \ - \ mg$$
Where $N$ is the normal force acting upwards on the block. We know that the impulse will be equal to the change in momentum, thus we have:
$$mv_0 \sin \theta \ = \ \displaystyle\int_{0}^{T} (N \ - \ mg) dt \ = \ \displaystyle\int_{0}^{T} N dt \ - \ mgT$$
Where $T$ is the amount of time it takes to reduce the block's vetical momentum to $0$.
Now, let's consider the momentum in the horizontal direction. Specifically, we wish to consider the change in momentum of the block between the point when it first comes into
contact with the ground, and $T$. We know that the only force acting in the horizontal direction is friction, which is given by $F_f \ = \ N\mu$, thus, if we denote the velocity
of the block at time $T$ by $v_T$, we can set up our "impulse relationship":
$$mv_0 \cos \theta\ - \ mv_T \ = \ \mu \displaystyle\int_{0}^{T} N dt$$
Now, we have two equation that we can work with! Specifically, we can rearrange the first equation and substitute it into the second:
$$\displaystyle\int_{0}^{T} N dt \ = \ mgT \ + \ mv_0 \sin \theta \ \Rightarrow \ mv_0 \cos \theta \ - \ mv_T \ = \ \mu mgT \ + \ \mu mv_0 \sin \theta$$
$$\Rightarrow \ v_T \ = \ v_0 \big[\cos \theta \ - \ \mu \sin \theta \big] \ - \ \mu gT$$
Now, we have arrived at a very important quantity: we know how fast the block is moving after its vertical momentum has been reduced to $0$. From this point on, we can determine normal force, and
thus the exact value of the force of friction very easily: since the ground no longer has to supply extra force to bring the block's vertical velocity to $0$, the only normal force it needs to supply
is that which balances the force force of gravity. Thus, for our new normal force $N'$, we have $N' \ = \ mg$. We can now take an energy-based approach to determine how far our block will travel while sliding
on the ground, by setting the initial kinetic energy equal to the work done by friction over some distance, which we will solve for:
$$N' D \mu \ = \ mg D \mu \ = \ \frac{1}{2} m v_T^2 \ = \ \frac{1}{2} m \big( v_0 \big[\cos \theta \ - \ \mu \sin \theta \big] \ - \ \mu gT \big)^2$$
$$\Rightarrow \ D \ = \ \frac{1}{2 \mu g} \big( v_0 \big[\cos \theta \ - \ \mu \sin \theta \big] \ - \ \mu gT \big)^2$$
Now there is one more component to this problem we must consider beforewe can find the optimal angle: the horizontal distance that the brick travels in the air. This can be determined
fairly easily. We find the time at which the brick hits the ground by solving the veritical equation of motion and setting $y \ = \ 0$:
$$v_0 \sin \theta t \ - \ \frac{1}{2} gt^2 \ = \ 0 \ \Rightarrow \ t \ = \ \frac{2 v_0 \sin \theta}{g}$$
We can then substitute this into the equation that gives us $x(t)$:
$$X \ = \ v_0 \cos \theta t \ = \ \frac{v_0^2 \sin 2\theta}{g}$$
Thus, the total range of the brick is given by:
$$R \ = \ X \ + \ D \ = \ \frac{v_0^2 \sin 2\theta}{g} \ + \ \frac{1}{2 \mu g} \big( v_0 \big[\cos \theta \ - \ \mu \sin \theta \big] \ - \ \mu gT \big)^2$$
Now, we can find the maximum range $R_{\text{Max}}$ by simply taking the derivative with respect to $\theta$, setting to $0$, and solving to finding the optimal angle! We will have:
$$\frac{d R}{d\theta} \ = \ \frac{2 v_0^2}{g} \cos 2\theta \ - \ \frac{1}{\mu g} \big( v_0 \big[\cos \theta \ - \ \mu \sin \theta \big] \ - \ \mu gT \big)
\big( v_0 \big[\sin \theta \ + \ \mu \cos \theta \big] \big)$$
Now, to simplify, Morin says that we should consider the situation in which the ground does not deform during the collision. This implies that the time over which
the collision takes place is approaching $0$, as in order to change velocity over a very very short distance, the force must be very large, which implies that the acceleration
is very large, which implies that for finite initial and final velocities, the time interval over which the force actsmust approach $0$. Thus, we have $T \ \rightarrow \ 0$, and we get:
$$2 \cos 2\theta \ = \ \frac{1}{\mu} \big[ \mu \cos^2 \theta \ - \ \mu \sin^2 \theta \ + \ \sin \theta \cos \theta \ - \ \mu^2 \sin \theta \cos \theta \big]$$
$$\Rightarrow \ 2 \mu \cos 2\theta \ = \ \mu \cos 2\theta \ + \ \frac{1}{2} \sin 2\theta (1 \ - \ \mu^2) \ \Rightarrow \ 2 \mu \cos 2\theta \ = \ \sin 2\theta (1 \ - \ \mu^2)$$
Which gives us:
$$\frac{\sin 2\theta}{\cos 2\theta} \ = \ \frac{2 \sin \theta \cos \theta}{\cos^2 \theta \ - \ \sin^2 \theta} \ = \ \frac{2 \tan \theta}{1 \ - \ \tan^2 \theta} \ = \ \frac{2 \mu}{1 \ - \ \mu^2}$$
Solving this gives us:
$$\tan \theta \ = \ \mu$$
Which in turn gives us the optimal angle at which we should launch our brick to travel the longest distance.