Question

A firefighting crew uses a water cannon that shoots water at 25.0 m/s at a fixed angle of 47.0 ∘ above the horizontal. The firefighters want to direct the water at a blaze that is 15.0 m above ground level.

Answer 1

$$y=v_{0y}t-0.5gt^2=v_{0}\sin{47\degree}t-0.5gt^2$$

$$15=25\sin{47\degree}t-0.5(9.8)t^2$$

$$t_1=1.218\ s, t_2=2.513\ s$$

$$d_1=25\cos{47\degree}(1.218)=20.8\ m$$

$$d_2=25\cos{47\degree}(2.513)=42.9\ m$$