The Arlington fire department is using a large tank of diameter 2D and height D.

Question

The Arlington fire department is using a large tank of diameter 2D and height D. The fire hose will shoot water at 45 degrees above the horizontal. The front of the tank is a distance of 6D from the fire hose. What is the minimum initial speed (v_0) of the water such that it enters the tank? What is the maximum initial speed (v_0) of the water such that it enters the tank Now assume the Arlington fire department is a distance of 8D from the tank. What is the minimum initial speed (v_0) of the water such that it enters the tank? What is the maximum initial speed (v_0) of the water such that it enters the tank Give your answer in terms of D and g (g = the acceleration of gravity)

Explanation / Answer

A)

Consider the Motion along X-axis :

X = 6D

Vox = Vo cos45

X = Vox t

t = 6D/(Vo cos45)                                    eq-1

consider the motion along Y-direction :

Y = 2D

Voy = Vo Sin45

a = - g

using the equation

Y = Voy t + (0.5) a t2

2D = (Vo Sin45) (6D/(Vo cos45)) - (0.5)g (6D/(Vo cos45))2

Vo = sqrt(9Dg)

B)

Consider the Motion along X-axis :

X = 8D

Vox = Vo cos45

X = Vox t

t = 8D/(Vo cos45)                                    eq-1

consider the motion along Y-direction :

Y = 2D

Voy = Vo Sin45

a = - g

using the equation

Y = Voy t + (0.5) a t2

2D = (Vo Sin45) (8D/(Vo cos45)) - (0.5)g (8D/(Vo cos45))2

Vo = 3.3 sqrt(Dg)

c)

Consider the Motion along X-axis :

X = 8D

Vox = Vo cos45

X = Vox t

t = 8D/(Vo cos45)                                    eq-1

consider the motion along Y-direction :

Y = 2D

Voy = Vo Sin45

a = - g

using the equation

Y = Voy t + (0.5) a t2

2D = (Vo Sin45) (8D/(Vo cos45)) - (0.5)g (8D/(Vo cos45))2

Vo = 3.3 sqrt(Dg)

d)

Consider the Motion along X-axis :

X = 10D

Vox = Vo cos45

X = Vox t

t = 10D/(Vo cos45)                                    eq-1

consider the motion along Y-direction :

Y = 2D

Voy = Vo Sin45

a = - g

using the equation

Y = Voy t + (0.5) a t2

2D = (Vo Sin45) (10D/(Vo cos45)) - (0.5)g (10D/(Vo cos45))2

Vo = 3.54 sqrt(Dg)

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