Question

A gardener uses a water hose to fill a 30.0 L bucket. The gardener notes that it takes 1.00 min to fill the bucket. A nozzle with an opening of cross sectional area 0.500 cm2 is then attached to the hose. The nozzle is held so that water is projected horizontally from a point 1.00 m above the ground. Over what horizontal distance can the water be projected

Answer 1

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:

$t=\sqrt{\frac{-2y_{f} }{g} }$ (eq. 2)

The equation for position is:

xf = xi + vxt = 0 + v2t = v2t (eq. 3)

Replacing equation 1 and 2 in equation 3:

$x_{f} =\frac{R}{A_{2} } \sqrt{\frac{-2y_{f} }{g} } =\frac{30}{0.5} \sqrt{\frac{(-2)*(-1)}{9.8} }*\frac{1000}{60} =451.75cm=4.51m$