Two children, Ferdinand and Isabella, are playing with a water hose on a sunny summer day. Isabella is holding the hose in her hand 1.0 meters above the ground and is trying to spray Ferdinand, who is standing 10.0 meters away. To increase the range of the water, Isabella places her thumb on the hose hole and partially covers it. Assuming that the flow remains steady, what fraction Image for Two children, Ferdinand and Isabella, are playing with a waterhose on a sunny summer day. Isabella is holding of the cross-sectional area of the hose hole does shehave to cover to be able to spray her friend?

 f = 0.84 or 84%

Step-by-step explanation:

Given:

- Original constant velocity of water v_o = 3.5 m/s

- The original diameter of the hose opening d_o = 1.5 cm

- The total vertical distance y(f) = 1.0 m

- The total range x(f) = 10.0 m

Find:

What fraction of the original area does Isabella has to reduce or pinch so the water can cover the entire range.

Solution:

- We will assume the flow of water as a stream of particles that follow a path of single point.

- Using second kinematic equation in the vertical direction and compute the time it takes the water to reach the ground:

                                  y(t) = y(0) + v_y*t + 0.5*gt^2

Where, y(0) = v_y = 0,

                                  1.0 = 0.5*g*t^2

                                  t = sqrt ( 2*1 / 9.81 )

                                  t = 0.4515 s

- Now we will use the second equation of motion for the x- direction motion:

                                 x(t) = x(0) + v_x*t

Where, x(0) = 0,

                                 10 = 0 + v_x*(0.4515)

                                 v_x = 10 / 0.4515 = 22.1 m/s

- We have calculated the minimum velocity required to reach the range from isabella and ferdinand.

- We will use the continuity equation to compute the area required for the velocity calculated v_x = 22.1 m/s:

                                A_f*v_x = A_o*v_o

                                A_f = A_o*v_o / v_x

                                A_f = pi*0.015^2*3.5 / 4*22.1

                                A_f = 1.12 * 10^-4 m^2

- The fraction f of the reduction of area is:

                                f = (A_o - A_f) / A_o

                                f = (7.068*10^-8 - 1.12*10^-4) / (7.068*10^-8)

                               f = 0.84 or 84%

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step-by-step explanation: