## Physics: Principles with Applications (7th Edition)

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.