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?

AThe 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.

johanrusli johanrusli · Answer 2 · 2019-08-17T07:15:31-04:00

(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.

$\large {\boxed {a = \frac{v - u}{t} } }$

$\large {\boxed {d = \frac{v + u}{2}~t } }$

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:

$v^2 = u^2 - 2gh$

$0^2 = u^2 - 2 \times 32.2 \times 560$

$u^2 = 36064$

$u = \sqrt{36064}$

$\large {\boxed {u \approx 190 ~ ft/s} }$

Question b:

$v = u - gt$

$0 = \sqrt{36064} - 32.2t$

$t = \sqrt{36064} \div 32.2$

$\large {\boxed {t \approx 5.90 ~ s} }$

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