Answer
They should position their cannon a distance of 8.77 meters or 52.7 meters from the building.
Work Step by Step
We can find the initial vertical velocity $v_{0y}$ of the water.
$v_{0y} = v_0~sin(\theta) = (25.0~m/s)~sin(53.0^{\circ})$
$v_{0y} = 20.0~m/s$
We can find the times $t$ when the water is 10.0 meters above the ground.
$y = v_{0y}~t - \frac{1}{2}gt^2$
$10.0~m = (20.0~m/s)~t - (4.90~m/s^2)~t^2$
$(4.90~m/s^2)~t^2 - (20.0~m/s)~t +10.0~m = 0$
We can use the quadratic formula to find $t$.
$t = \frac{-b\pm \sqrt{b^2-4ac}}{2a}$
$t = \frac{-(20.0)\pm \sqrt{(-20.0)^2-(4)(4.90)(10.0)}}{(2)(4.90)}$
$t = 0.583~s, 3.50~s$
We can use these two values of $t$ to find the distance $x$ from the building.
$x_1 = v_x~t = (25.0~m/s)~cos(53.0^{\circ})(0.583~s)$
$x_1 = 8.77~m$
$x_2 = v_x~t = (25.0~m/s)~cos(53.0^{\circ})(3.50~s)$
$x_2 = 52.7~m$
They should position their cannon a distance of 8.77 meters or 52.7 meters from the building.
Note that when the distance is 8.77 meters from the building, the water reaches 10.0 meters high on the building while the water is on its way up. When the distance is 52.7 meters from the building, the water reaches 10.0 meters high on the building after the water has gone up to its maximum height and the water is on its way down.
