Answer
$\textbf{(a)}$ $l=29.3\text{ m}$ away from the base of the monument.
$\textbf{(b)}$ The speed is $v=57.8\text{ m/s}$ directed $85^\circ$ below the horizontal.
Work Step by Step
The height of the monument, transformed to SI units, is $h=169\text{ m}$. (one ft is 30.48m).
$\textbf{(a)}$
Since the ball is thrown horizontally, the time of the flight is given by
$$h=\frac{1}{2}gt^2\Rightarrow t=\sqrt{\frac{2h}{g}}.$$
Now, in the horizontal direction, the component of the velocity is constant so we have
$$l=v_h t=v_h\sqrt{\frac{2h}{g}}=5.00\text{ m/s}\sqrt{\frac{2\cdot169\text{ m}}{9.81\text{ m/s}^2}}=29.3\text{ m},$$
i.e. the ball landed $30\text{ m}$ from the base of the monument in the direction of throwing.
$\textbf{(b)}$
Since the initial vertical component of the velocity is equal to zero, the vertical component of the velocity at the landing is given by
$$v_v=\sqrt{2gh}.$$
By the pythagorean theorem, we have for the magntude of the velocity
$$v=\sqrt{v_h^2+v_v^2}=\sqrt{v_h^2+2gh}=\sqrt{5.00^2+2\cdot9.81\cdot169}\text{ m/s}=57.8\text{ m/s}.$$
The vertical and the horizontal component make the leggs of the right triangle so we have for the angle:
$$\tan\theta=\frac{v_v}{v_h}=\frac{\sqrt{2gh}}{v_h}$$
which gives
$$\theta=\arctan\left(\frac{\sqrt{2gh}}{v_h}\right)=\arctan\left(\frac{\sqrt{2\cdot9.81\text{m/s}^2\cdot169\text{ m}}}{5.00\text{ m}}\right)=85^\circ,$$
so the velocity is directed $85^\circ$ below the horizontal.
