Question

Two children, Ferdinand and Isabella, are playing with a waterhose on a sunny summer day. Isabella is holding the hose in herhand 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 f of the cross-sectional area of the hose hole does she have to cover to be able to spray her friend? Assume that the cross section of the hose opening is circular with a radius of 1.5 centimeters. Express your answer as a percentage to the nearest integer.

Answer 1

Answer:

84%

Explanation:

g = Acceleration due to gravity = 9.81 m/s²

$y_0=1\ m$

x = 10 m

t = Time taken

$v_0$ = 3.5 m/s (assumed, as it is not given)

$A_0=\pi 1.5^2$

We have the equation

$y=y_0+ut+\dfrac{1}{2}gt^2\\\Rightarrow 0=y_0-\dfrac{1}{2}gt^2\\\Rightarrow t=\sqrt{\dfrac{2y_0}{g}}\\\Rightarrow t=\sqrt{\dfrac{2\times 1}{9.81}}\\\Rightarrow t=0.45152\ s$

$x=x_0+vt\\\Rightarrow 10=0+v0.45152\\\Rightarrow v=\dfrac{10}{0.45152}\\\Rightarrow v=22.14741\ m/s$

From continuity equation we have

$Av=A_0v_0\\\Rightarrow A=\dfrac{A_0v_0}{v}\\\Rightarrow A=\dfrac{\pi 1.5^2\times 3.5}{22.14741}\\\Rightarrow A=1.11706\ cm^2$

Fraction is given by

$f=\dfrac{A_0-A}{A_0}\times 100\\\Rightarrow f=\dfrac{\pi 1.5^2-1.11706}{\pi 1.5^2}\times 100\\\Rightarrow f=84.196\ \%\approx 84\ \%$

The fraction is 84%