Projectile motion question A projectile is fired with an initial velocity of $V$ inclined at an angle alpha above the horizontal.
Assuming that the only involved force is the gravity, write differential equations that describe the trajectory $(x(t), y(t))$ and solve them to compute $x(t)$ and $y(t)$.
The projectile is first fired horizontally from a point $O$ which is at the top of a cliff, so as to hit a fixed target in the water, and it is observed that the time of the flight is $T$.
It is found that, with the same initial speed, the target can be hit by firing at an angle alpha above the horizontal.
Show that the distance of the target from the point at sea level
vertically below $O$ is equal to $\displaystyle \frac{1}{2}*g*T^2*\tan(\alpha)$
 A: Assuming you have solved the differential equations, the equations of motion are
$$x(t) = v_0 \cos{\alpha} t$$
$$y(t) = y_0 + v_0 \sin{\alpha} t - \frac{1}{2} g t^2$$
Note that $y_0$ is the unknown height of the cliff.
Now, the initial condition specifies that the target is hit when $\alpha=0$.  In that case, 
$$y(T) = 0 = y_0-\frac{1}{2} g T^2 \implies y_0 = \frac{1}{2} g T^2$$
Also, 
$$x(T) = D = v_0 T$$
where $D$ is the distance from the target to the foot of the cliff.
Now, for $\alpha$ nonzero in the second condition, we may combine the first two equations to get
$$y(x) = y_0 + \tan{\alpha} x - \frac{g \sec^2{\alpha}}{2 v_0^2} x^2$$
Again, using $y(D)=0$ and substituting the relations we found in the horizontal case, we get
$$\frac{1}{2} g T^2 + D\tan{\alpha}  - \frac{g \sec^2{\alpha}}{2 v_0^2}D^2=0$$
$$\implies \frac{1}{2} g T^2 + D\tan{\alpha}  - \frac{g \sec^2{\alpha}}{2}T^2=0$$
Use the fact that $\sec^2{\alpha} = 1 + \tan^2{\alpha}$ and get
$$ D \tan{\alpha} - \frac{1}{2} g T^2 \tan^2{\alpha} = 0$$
from which the desired result follows.
