Question

Water in a 6cm diameter pipe is moving at a speed of 60m/s. How fast is it moving if the pipe is reduced to a .05cm opening? If the pipe is 4m off the ground how far in front of the pipe will the water land?

Answer 1

Using Continuity equation:

v1*A1 = v2*A2

v1*d1^2 = v2*d2^2

v1 = 60 m/sec

d1 = 6 cm

d2 = 0.05 cm

v2 = v1*(d1^2/d2^2)

v2 = 60*(6^2/0.05^2) = 8.64*10^5 m/sec

Now

Vertical height = 4 m/sec

Viy = 0 m/sec

Vix = 8.64*10^5 m/sec

ay = -9.8 m/sec^2

ax = 0 m/sec^2

In y-direction using the equation

h = Viy*t + 0.5*a*t^2

-4 = 0*t - 0.5*9.81*t^2

t = sqrt (4*2/9.81) = 0.903 sec

Now distance traveled in x-direction will be

x = Vix*t

x = 8.64*10^5*0.903 = 7.8*10^5 m

Let me know if you have any doubt.