Answer
See answers.
Work Step by Step
a. The water starts and ends at ground level. Use the projectile motion range equation.
$$R=\frac{v_0^2sin 2 \theta}{g}$$
$$v_0=\sqrt{\frac{Rg}{sin 2 \theta}}=\sqrt{\frac{(6.0m)(9.80m/s^2)}{sin 70^{\circ}}}=7.9m/s$$
b. The volume rate of flow is the area of the flow, multiplied by the speed of the flow. There are four sprinkler heads.
$$Flow=Av=4\pi r^2v=4 \pi (0.0015m)^2(7.910m/s)$$
$$=2.237\times10^{-4}m^3/s=0.22\;liters/s$$
c. Apply the equation of continuity.
$$(Av)_{supply}=(Av)_{heads}$$
Calculate the flow speed in the pipe.
$$v_{supply}=\frac{(Av)_{heads}}{A_{supply}}$$
$$v_{supply}=\frac{2.237\times10^{-4}m^3/s }{\pi (0.0095m)^2}=0.79m/s$$