Answer:
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 = [tex]\sqrt{(-v0^{2} * sin \alpha * cos\alpha / g)^{2}}[/tex]
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
