## College Physics (4th Edition)

We can find the time it takes for the "projectile" to reach a vertical displacement of $-5.0~m$: $\Delta y = v_{0y}~t+\frac{1}{2}a_yt^2$ $-5.0 = (18.0)~sin~35.0^{\circ}~t+\frac{1}{2}(-9.80)t^2$ $4.9~t^2-10.32~t-5.0 = 0$ We can use the quadratic formula to find $t$: $t = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$ $t = \frac{10.32 \pm\sqrt{(-10.32)^2 - 4(4.9)(-5.0)}}{(2)(4.9)}$ $t = -0.406~s, 2.51~s$ Since the time must be positive, it takes 2.51 seconds for the "projectile" to reach a vertical displacement of $-5.0~m$. We can find the horizontal displacement at this time: $\Delta x = v_x~t = (18.0~m/s)~cos~35.0^{\circ}~(2.51~s) = 37.0~m$ The net should be placed a horizontal distance of 37.0 meters away from the cannon.