Water leaves a fireman’s hose (held near the ground) with an initial velocity ${v_0 = 22.5 m/s}$ at an angle θ = 35° above horizontal. Assume the water acts as a projectile that moves without air resistance.
From the building base, where should the fireman place the hose for the water to reach its maximum height as it strikes the building? Express this distance, d, in terms of ${v_0}$, θ, and g.
Okay so can someone explain why it is not $${v_o^2\sin(2\theta)/g}$$
Best Answer
The formula ${v_o^2\sin(2\theta)/g}$ tells us how far away the water will hit the ground if the hose is on a large flat plane.
That would be a good distance at which to place the hose if you want the water to just barely hit the base of the building (where the building meets the ground). But you want the water to be higher when it hits, so you must stand closer.
If you draw a diagram of what is happening, it may be more obvious how far from the building the hose should be.