#### Answer

The rock takes 3.1 seconds to hit the ground.

#### Work Step by Step

We first write an expression for the horizontal displacement during the time of flight $t$:
$x = v_x ~t$
We then write an expression for the vertical displacement during the time of flight $t$:
$y = \frac{1}{2}at^2$
Since the slope is at a 45-degree angle, the horizontal displacement must equal the downward vertical displacement:
$y = x$
$\frac{1}{2}at^2 = v_x ~t$
$\frac{1}{2}(9.80 ~m/s^2) ~t^2 = (15 ~m/s) ~t$
$t = \frac{(2)(15 ~m/s)}{9.80 ~m/s^2} = 3.1 ~s$
The rock takes 3.1 seconds to hit the ground.