Water leaves a fireman’s hose (held near the ground) with an initial velocity v0= 22.5 m/s at an angle θ = 28.5° above horizontal. assume the water acts as a projectile that moves without air resistance. use a cartesian coordinate system with the origin at the hose nozzle position.
a) using v0, θ, and g, write an expression for the time, tmax, the water travels to reach its maximum vertical height.
b) at what horizontal distance d 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 v0, θ, and g.

a) tmax = -v0 sin θ / g

b) d = - v0² sin (2θ)  / 2g

Explanation:

Please, see the attached figure for a graphical description of the problem.

The water moves following a parabolic motion.

a) The vector velocity (v) at time t is given by the following expression:

v = (v0 cos θ ; v0 sin θ + g t)

where

v0 = initial speed

θ = angle at which the water leaves the hose

g = acceleration due to gravity (9.8 m/s²)

t = time

When the water reaches its maximum vertical height, the y-component of the velocity vector equals 0, since the water, for that instance, does not go up nor down. Then:

v0 sin θ + g tmax = 0

tmax = -v0 sin θ / g

tmax = -22.5 m/s * sin 28.5° / (-9.8 m/s²) = 1.1 s

b) The vector position at time t is written as follows (see figure):

r = (x0 + v0 t cos θ; y0 + v0 t sin θ + 1/2 g t²)

since the hose is at the origin, x0 = 0 and y0 = 0.

r = ( v0 t cos θ;  v0 t sin θ + 1/2 g t²)

The distance is the module of the rx vector (see figure), which is the x-component of the r  vector a t = t max:

rx = (v0 tmax cos θ; 0)

Replacing t max = -v0 sin θ / g

rx = (- v0² sin θ cos θ / g ; 0)

d = module rx =

d = - v0² sin θ cos θ / g

Using trigonometric identity: sin x cos x = sin (2x) / 2

d = - v0² sin (2θ)  / 2g

d = (22.5 m/s)² sin (2* 28.5) / 2* 9.8 m/s² = 21.  7 m

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