Answer:
The distance can the water be projected is 4.51 m
Explanation:
The speed of the water in the hose is equal to:
v1 = R/A1
If we solve the continuity for v2:
v2 = R/A2 (eq. 1)
The equation for the vertical position is:
yf = yi + vy*t - (1/2)gt²
yi = 0
vy = 0
Clearing t:
[tex]t=\sqrt{\frac{-2y_{f} }{g} }[/tex] (eq. 2)
The equation for position is:
xf = xi + vxt = 0 + v2t = v2t (eq. 3)
Replacing equation 1 and 2 in equation 3:
[tex]x_{f} =\frac{R}{A_{2} } \sqrt{\frac{-2y_{f} }{g} } =\frac{30}{0.5} \sqrt{\frac{(-2)*(-1)}{9.8} }*\frac{1000}{60} =451.75cm=4.51m[/tex]
