The Physics of the Problem

Oct 13, 2025 by thesweetdesigns.com 30 views

Introduction

The world of physics is full of fascinating phenomena that can be modeled using mathematical equations. One such phenomenon is the motion of an object under the influence of gravity. In this article, we will explore the motion of a cliff diver jumping from a height of 105 feet with an initial upward velocity of 5 feet per second. We will derive an equation that models the height of the diver above the water in time elapsed.

The Physics of the Problem

When the diver jumps from the cliff, he experiences an upward force due to his initial velocity. However, as he descends, he is also subject to the downward force of gravity. The acceleration due to gravity is a constant 32 feet per second squared. We can use the following kinematic equation to model the height of the diver:

h(t) = h_0 + v_0t - \frac{1}{2}gt^2

where:

$h(t)$ is the height of the diver above the water at time $t$
$h_0$ is the initial height of the diver (105 feet)
$v_0$ is the initial upward velocity of the diver (5 feet per second)
$g$ is the acceleration due to gravity (32 feet per second squared)
$t$ is the time elapsed in seconds

Deriving the Equation

To derive the equation, we can start by considering the motion of the diver in the vertical direction. Since the diver is subject to the downward force of gravity, his velocity will decrease as he descends. We can use the following kinematic equation to model the velocity of the diver:

v(t) = v_0 - gt

where:

$v(t)$ is the velocity of the diver at time $t$
$v_0$ is the initial upward velocity of the diver (5 feet per second)
$g$ is the acceleration due to gravity (32 feet per second squared)
$t$ is the time elapsed in seconds

Now, we can use the following kinematic equation to model the height of the diver:

h(t) = h_0 + v_0t - \frac{1}{2}gt^2

where:

$h(t)$ is the height of the diver above the water at time $t$
$h_0$ is the initial height of the diver (105 feet)
$v_0$ is the initial upward velocity of the diver (5 feet per second)
$g$ is the acceleration due to gravity (32 feet per second squared)
$t$ is the time elapsed in seconds

Solving the Equation

To solve the equation, we can plug in the values of the variables:

h(t) = 105 + 5t - \frac{1}{2}(32)t^2

Simplifying the equation, we get:

h(t) = 105 + 5t - 16t^2

This is the equation that models the height of the diver above the water in time elapsed.

Graphing the Equation

To visualize the motion of the diver, we can graph the equation:

import numpy as np
import matplotlib.pyplot as plt

# Define the equation
def h(t):
    return 105 + 5*t - 16*t**2

# Generate values of t
t = np.linspace(0, 10, 100)

# Calculate corresponding values of h
h_values = h(t)

# Plot the graph
plt.plot(t, h_values)
plt.xlabel('Time (s)')
plt.ylabel('Height (ft)')
plt.title('Height of Diver above Water')
plt.grid(True)
plt.show()

This code generates a graph of the height of the diver above the water in time elapsed.

Conclusion

In this article, we have derived an equation that models the height of a cliff diver above the water in time elapsed. We have used the kinematic equations to model the motion of the diver and have solved the equation to obtain a quadratic function. We have also graphed the equation to visualize the motion of the diver. This equation can be used to predict the height of the diver at any given time, which can be useful in various applications such as skydiving and parachuting.

References

Halliday, D., Resnick, R., & Walker, J. (2013). Fundamentals of Physics. John Wiley & Sons.
Serway, R. A., & Jewett, J. W. (2018). Physics for Scientists and Engineers. Cengage Learning.

Introduction

In our previous article, we explored the motion of a cliff diver jumping from a height of 105 feet with an initial upward velocity of 5 feet per second. We derived an equation that models the height of the diver above the water in time elapsed. In this article, we will answer some frequently asked questions about the motion of the cliff diver.

Q: What is the initial height of the cliff diver?

A: The initial height of the cliff diver is 105 feet.

Q: What is the initial upward velocity of the cliff diver?

A: The initial upward velocity of the cliff diver is 5 feet per second.

Q: What is the acceleration due to gravity?

A: The acceleration due to gravity is 32 feet per second squared.

Q: How can I use the equation to predict the height of the cliff diver at any given time?

A: To predict the height of the cliff diver at any given time, you can plug in the value of time into the equation:

h(t) = 105 + 5t - 16t^2

For example, if you want to know the height of the cliff diver at 5 seconds, you can plug in t = 5 into the equation:

h(5) = 105 + 5(5) - 16(5)^2

Simplifying the equation, you get:

h(5) = 105 + 25 - 400

h(5) = -270

So, the height of the cliff diver at 5 seconds is -270 feet.

Q: What is the maximum height reached by the cliff diver?

A: To find the maximum height reached by the cliff diver, we need to find the time at which the velocity of the cliff diver is zero. We can do this by setting the velocity equation equal to zero and solving for time:

v(t) = v_0 - gt

0 = 5 - 32t

Solving for t, we get:

t = \frac{5}{32}

Now, we can plug this value of t into the height equation to find the maximum height reached by the cliff diver:

h(t) = 105 + 5t - 16t^2

h(\frac{5}{32}) = 105 + 5(\frac{5}{32}) - 16(\frac{5}{32})^2

Simplifying the equation, we get:

h(\frac{5}{32}) = 105 + \frac{25}{32} - \frac{25}{64}

h(\frac{5}{32}) = 105 + \frac{25}{32} - \frac{25}{64}

h(\frac{5}{32}) = 105 + \frac{50}{64} - \frac{25}{64}

h(\frac{5}{32}) = 105 + \frac{25}{64}

h(\frac{5}{32}) = 105 + 0.390625

h(\frac{5}{32}) = 105.390625

So, the maximum height reached by the cliff diver is approximately 105.39 feet.

Q: What is the time it takes for the cliff diver to hit the water?

A: To find the time it takes for the cliff diver to hit the water, we need to find the time at which the height of the cliff diver is zero. We can do this by setting the height equation equal to zero and solving for time:

h(t) = 105 + 5t - 16t^2

0 = 105 + 5t - 16t^2

Rearranging the equation, we get:

16t^2 - 5t - 105 = 0

Solving for t, we get:

t = \frac{5 \pm \sqrt{25 + 6720}}{32}

t = \frac{5 \pm \sqrt{6745}}{32}

t = \frac{5 \pm 82.03}{32}

Since time cannot be negative, we take the positive root:

t = \frac{5 + 82.03}{32}

t = \frac{87.03}{32}

t = 2.72

So, the time it takes for the cliff diver to hit the water is approximately 2.72 seconds.

Conclusion

In this article, we have answered some frequently asked questions about the motion of a cliff diver jumping from a height of 105 feet with an initial upward velocity of 5 feet per second. We have used the equation to predict the height of the cliff diver at any given time, found the maximum height reached by the cliff diver, and determined the time it takes for the cliff diver to hit the water.