Since we don't know the volume of the ball, we may approximate it as a point particle. Some mention has been made of the friction between the ground and the ball - all I will say is that we should ignore it because a) the moment of inertia (or some way to find it) aren't specified, and b) if the ball did start spinning, the friction would become static, not kinetic. Still a doable problem, but you'd need the moment of inertia.
$$
\begin{eqnarray}
U_0 &=& \frac{m \dot x_0^2}{2} + m g h = 1490\text J\\
U_n &=& \frac{m \dot x_0^2}{2} + m g h - n
\end{eqnarray}$$
Where $m$ is the ball's mass, $v_0$ is the initial (horizontal) velocity, $g$ is the local surface gravity (assumed to be $9.8 \frac{\text m}{ \text s}$), $h$ is the initial height, and $n$ is the number of bounces and has units of Joules. Obviously, we must devise a sum over the energy from 1489 to 1 and add the duration of the first half parabola.
Even though the collision isn't perfectly elastic, I can think of no preference for a shallower or steeper angle after the bounce. Incidentally, that angle is $35^\circ$, and upon reflection, the ratio $\frac{\dot y_n}{\dot x_n} = \frac{\sqrt{2 g h}}{\dot x_0} = \eta \approx \frac{7}{10}$ would then be constant.
The duration of the $n$th bounce is $t = \frac{2 \dot y_n}{g}$ (from kinematics), so all we must do is put $y_n$ in terms of $U_n$. When the ball hits the ground, it only has kinetic energy.
$$
\begin{eqnarray}
U_n &=& \frac{m(\dot x_n^2 + \dot y_n^2)}{2} \\
U_n &=& \frac{m(\frac{\dot y_n^2}{\eta^2} + \dot y_n^2)}{2} \\
U_n &=& \frac{m\dot y_n^2}{2}\left( 1 + \frac{1}{\eta^2}\right) \\
\dot y_n &=& \sqrt{\frac{2 U_n}{m \left(1 + \frac{1}{\eta^2} \right)}}
\end{eqnarray}
$$
Doing a little algebra and writing the sum,
$$t_{total} = \sqrt{\frac{2 h}{g}} + \frac{2 \eta}{g} \sqrt{\frac{2}{m\left(1 + \eta^2\right)}} \sum_{n=1489}^1 \sqrt{U_n} \approx 1471 \text s$$
If you were writing a test, the sum of square roots is a formidable problem, but you can avoid that (using the apocryphal Gauss trick) if you ask for the distance traveled. I leave it as a (quite fun) exercise for the reader.