Answer
The shot would travel a horizontal distance of $~~24.9~m$
Work Step by Step
We can use the vertical components to find the time of flight:
$y = y_0+v_{y0}~t+\frac{1}{2}a_yt^2$
$0 = 2.160+(15.00~sin~45.00^{\circ})~t+\frac{1}{2}(-9.8)t^2$
$0 = 2.160+10.6~t-4.9t^2$
$4.9t^2-10.6~t-2.160 = 0$
We can use the quadratic formula:
$t = \frac{-(-10.6) \pm \sqrt{(-10.6)^2-(4)(4.9)(-2.160)})}{(2)(4.9)}$
$t = \frac{10.6 \pm \sqrt{154.696}}{9.8}$
$t = -0.188~s, 2.35~s$
Since $t$ is positive, we can use the positive solution.
We can find the horizontal distance the shot would travel:
$x = v_x~t$
$x = (15.00~m/s)(cos~45.00^{\circ})(2.35~s)$
$x = 24.9~m$
The shot would travel a horizontal distance of $~~24.9~m$
