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.
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
- Velocity of Runner : https://brainly.com/question/3813437
- Kinetic Energy : https://brainly.com/question/692781
- Acceleration : https://brainly.com/question/2283922
- The Speed of Car : https://brainly.com/question/568302
Answer details
Grade: High School
Subject: Physics
Chapter: Kinematics
Keywords: Velocity , Driver , Car , Deceleration , Acceleration , Obstacle , Speed , Time , Rate
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