Modeling the Motion of the Egg

Oct 13, 2025 by orovajewels.com 53 views

You are tossing an egg from the top of a building that is 100 feet above the ground. The egg will be thrown straight up at a velocity of 40 feet per second. Which function below represents the height $h$ of the egg after $t$ seconds?

When an object is thrown upwards, it experiences a downward acceleration due to gravity. The acceleration due to gravity is 32 feet per second squared (ft/s^2). We can use the kinematic equations to model the motion of the egg.

Kinematic Equations

The kinematic equations are a set of equations that describe the motion of an object under constant acceleration. The equations are:

$h = h_0 + v_0t - \frac{1}{2}gt^2$
$v = v_0 - gt$
$v^2 = v_0^2 - 2a(h - h_0)$

where:

$h$ is the height of the object at time $t$
$h_0$ is the initial height of the object
$v_0$ is the initial velocity of the object
$g$ is the acceleration due to gravity
$t$ is time

Modeling the Motion of the Egg

We are given that the egg is thrown straight up at a velocity of 40 ft/s from a height of 100 ft. We want to find the height of the egg after $t$ seconds.

Let's use the first kinematic equation to model the motion of the egg:

$h = h_0 + v_0t - \frac{1}{2}gt^2$

We know that $h_0 = 100$ ft, $v_0 = 40$ ft/s, and $g = 32$ ft/s^2. Plugging these values into the equation, we get:

$h = 100 + 40t - \frac{1}{2}(32)t^2$

Simplifying the equation, we get:

$h = 100 + 40t - 16t^2$

Representing the Height as a Function

We want to represent the height of the egg as a function of time $t$ . We can do this by writing the equation in the form $h(t)$ :

$h(t) = 100 + 40t - 16t^2$

This is the function that represents the height of the egg after $t$ seconds.

Conclusion

Example Use Cases

Finding the maximum height: To find the maximum height of the egg, we need to find the time at which the egg reaches its maximum height. We can do this by setting the derivative of the height function equal to zero and solving for $t$ .
Finding the time of impact: To find the time at which the egg hits the ground, we need to set the height function equal to zero and solve for $t$ .

Derivative of the Height Function

To find the maximum height of the egg, we need to find the derivative of the height function:

$\frac{dh}{dt} = 40 - 32t$

Setting the derivative equal to zero, we get:

$40 - 32t = 0$

Solving for $t$ , we get:

$t = \frac{40}{32} = \frac{5}{4}$

This is the time at which the egg reaches its maximum height.

Time of Impact

To find the time at which the egg hits the ground, we need to set the height function equal to zero and solve for $t$ :

$100 + 40t - 16t^2 = 0$

Solving for $t$ , we get:

$t = \frac{-40 \pm \sqrt{1600 + 6400}}{-32}$

$t = \frac{-40 \pm \sqrt{8000}}{-32}$

$t = \frac{-40 \pm 89.44}{-32}$

$t = \frac{-40 + 89.44}{-32}$ or $t = \frac{-40 - 89.44}{-32}$

$t = \frac{49.44}{-32}$ or $t = \frac{-129.44}{-32}$

$t = -1.54$ or $t = 4.04$

Since time cannot be negative, we discard the negative solution. Therefore, the egg hits the ground after $t = 4.04$ seconds.

Conclusion

In this problem, we used the kinematic equations to model the motion of an egg thrown straight up from a height of 100 ft. We found that the height of the egg after $t$ seconds is represented by the function $h(t) = 100 + 40t - 16t^2$ . We also found the time at which the egg reaches its maximum height and the time at which the egg hits the ground.
You are tossing an egg from the top of a building that is 100 feet above the ground. The egg will be thrown straight up at a velocity of 40 feet per second. Which function below represents the height $h$ of the egg after $t$ seconds?

Q: What is the initial velocity of the egg?

A: The initial velocity of the egg is 40 ft/s.

Q: What is the acceleration due to gravity?

A: The acceleration due to gravity is 32 ft/s^2.

Q: What is the equation for the height of the egg after $t$ seconds?

A: The equation for the height of the egg after $t$ seconds is:

$h(t) = 100 + 40t - 16t^2$

Q: What is the maximum height of the egg?

A: To find the maximum height of the egg, we need to find the time at which the egg reaches its maximum height. We can do this by setting the derivative of the height function equal to zero and solving for $t$ .

Q: How do we find the time of impact?

A: To find the time at which the egg hits the ground, we need to set the height function equal to zero and solve for $t$ .

Q: What is the time of impact?

A: The time of impact is 4.04 seconds.

Q: What happens to the egg after it hits the ground?

A: After the egg hits the ground, it will no longer be in the air and will not be subject to the same forces that were acting on it while it was in the air.

Q: Can we use the same equation to model the motion of the egg if it is thrown downwards?

A: No, we cannot use the same equation to model the motion of the egg if it is thrown downwards. The equation we derived assumes that the egg is thrown upwards, and the forces acting on the egg are different when it is thrown downwards.

Q: How do we account for air resistance in the equation?

A: Air resistance is a complex phenomenon that can affect the motion of the egg. However, in this problem, we are assuming that air resistance is negligible, and we are only considering the effects of gravity.

Q: Can we use the same equation to model the motion of other objects?

A: Yes, we can use the same equation to model the motion of other objects that are subject to the same forces. However, we need to make sure that the initial conditions and the forces acting on the object are the same.

Q: What are some real-world applications of this problem?

A: This problem has many real-world applications, such as:

Projectile motion: The motion of the egg can be used to model the motion of projectiles, such as bullets or rockets.
Aerodynamics: The motion of the egg can be used to study the effects of air resistance on objects in motion.
Physics education: This problem can be used to teach students about kinematics and the motion of objects under constant acceleration.

Conclusion

In this article, we used the kinematic equations to model the motion of an egg thrown straight up from a height of 100 ft. We found that the height of the egg after $t$ seconds is represented by the function $h(t) = 100 + 40t - 16t^2$ . We also answered some common questions about the problem and discussed some real-world applications of the problem.