The world's highest fountain of water is located, appropriately enough, in Fountain Hills, Arizona. The fountain rises to a height of 560 ft. (a) What is the initial speed of the water? (b) How long does it take for the water to reach the top of the fountain?

Respuesta :

a. The relationship between the height and initial velocity is given in the equation,
                          d = ((Vi)² - (Vf)²) / 2g
Vf for this case is zero and g is equal to 32.2 ft/s². Substituting,
                            560 ft = (Vi)² / (2)(32.2)
The value of Vi is equal to 189.91 ft/s.
b. For the time, the equation is,
                            d = 0.5gt²
Substituting,
                           560 ft = 0.5(32.2)(t²)
The value of t is 5.90 s. 

(a) The initial speed of the water is 190 ft/s

(b) It takes 5.90 s for the water to reach the top of the fountain

Further explanation

Acceleration is rate of change of velocity.

[tex]\large {\boxed {a = \frac{v - u}{t} } }[/tex]

[tex]\large {\boxed {d = \frac{v + u}{2}~t } }[/tex]

a = acceleration (m / s²)v = final velocity (m / s)

u = initial velocity (m / s)

t = time taken (s)

d = distance (m)

Let us now tackle the problem!

Given:

h = 560 ft

g = 32.2 ft/s²

Unknown:

u = ?

Solution:

At the maximum height, the speed is 0 ft/s

Question a:

[tex]v^2 = u^2 - 2gh[/tex]

[tex]0^2 = u^2 - 2 \times 32.2 \times 560[/tex]

[tex]u^2 = 36064[/tex]

[tex]u = \sqrt{36064}[/tex]

[tex]\large {\boxed {u \approx 190 ~ ft/s} }[/tex]

Question b:

[tex]v = u - gt[/tex]

[tex]0 = \sqrt{36064} - 32.2t[/tex]

[tex]t = \sqrt{36064} \div 32.2[/tex]

[tex]\large {\boxed {t \approx 5.90 ~ s} }[/tex]

Learn more

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Answer details

Grade: High School

Subject: Physics

Chapter: Kinematics

Keywords: Velocity , Driver , Car , Deceleration , Acceleration , Obstacle , Speed , Time , Rate

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