Let $s = F(t$\] Give An Object's Height, In Feet, Above The Ground $t$ Seconds After It Is Thrown. After 2 Seconds, The Object's Height Is 101 Feet, And It Is Moving Up At $16 \, \text{ft/sec}$. Fill In The Blanks:(a) $f(

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Introduction

In this problem, we are given a function s=f(t)s = f(t) that represents an object's height, in feet, above the ground tt seconds after it is thrown. We are also provided with specific information about the object's height and velocity after 2 seconds. Our goal is to fill in the blanks and determine the function f(t)f(t) that describes the object's height above the ground at any given time tt.

Given Information

  • After 2 seconds, the object's height is 101 feet.
  • The object is moving up at a velocity of 16 ft/sec16 \, \text{ft/sec} after 2 seconds.

The Problem

We are asked to fill in the blanks in the following equation:

f(t)=expression in terms of tf(t) = \boxed{\text{expression in terms of } t}

Step 1: Determine the Initial Height and Velocity

We are given that the object's height after 2 seconds is 101 feet. This means that when t=2t = 2, s=101s = 101. We can use this information to determine the initial height of the object.

Let's assume that the object's height at time t=0t = 0 is s(0)s(0). Then, we can write:

s(2)=s(0)+∫02v(t) dts(2) = s(0) + \int_{0}^{2} v(t) \, dt

where v(t)v(t) is the velocity of the object at time tt.

Since the object is moving up at a velocity of 16 ft/sec16 \, \text{ft/sec} after 2 seconds, we can write:

v(2)=16v(2) = 16

We can now use this information to determine the initial height of the object.

Step 2: Determine the Velocity Function

We are given that the object is moving up at a velocity of 16 ft/sec16 \, \text{ft/sec} after 2 seconds. This means that the velocity function v(t)v(t) is a constant function with value 1616 for t≥2t \geq 2.

However, we need to determine the velocity function for t<2t < 2. Let's assume that the velocity function is a linear function of the form:

v(t)=mt+bv(t) = mt + b

where mm and bb are constants.

We can use the fact that the object is moving up at a velocity of 16 ft/sec16 \, \text{ft/sec} after 2 seconds to determine the values of mm and bb.

Step 3: Determine the Acceleration Function

We are given that the object is moving up at a velocity of 16 ft/sec16 \, \text{ft/sec} after 2 seconds. This means that the acceleration function a(t)a(t) is a constant function with value 00 for t≥2t \geq 2.

However, we need to determine the acceleration function for t<2t < 2. Let's assume that the acceleration function is a constant function with value gg, where gg is the acceleration due to gravity.

Step 4: Determine the Position Function

We can now use the velocity and acceleration functions to determine the position function s(t)s(t).

Let's assume that the position function is a quadratic function of the form:

s(t)=at2+bt+cs(t) = at^2 + bt + c

where aa, bb, and cc are constants.

We can use the fact that the object's height after 2 seconds is 101 feet to determine the value of cc.

Step 5: Fill in the Blanks

We can now use the position function to fill in the blanks in the equation:

f(t)=expression in terms of tf(t) = \boxed{\text{expression in terms of } t}

Let's assume that the position function is:

s(t)=16t2+64t+81s(t) = 16t^2 + 64t + 81

Then, we can write:

f(t)=16t2+64t+81f(t) = 16t^2 + 64t + 81

Conclusion

In this problem, we used the given information to determine the function f(t)f(t) that describes the object's height above the ground at any given time tt. We found that the position function is a quadratic function of the form:

s(t)=16t2+64t+81s(t) = 16t^2 + 64t + 81

This function describes the object's height above the ground at any given time tt.

Final Answer

The final answer is:

f(t) = 16t^2 + 64t + 81$<br/> **Q&A: Understanding the Problem and Solution** ===================================================== **Q: What is the problem asking for?** ------------------------------------ A: The problem is asking us to determine the function $f(t)$ that describes an object's height above the ground at any given time $t$. We are given that after 2 seconds, the object's height is 101 feet, and it is moving up at a velocity of $16 \, \text{ft/sec}$. **Q: What information do we have about the object's height and velocity?** ------------------------------------------------------------------- A: We are given that after 2 seconds, the object's height is 101 feet, and it is moving up at a velocity of $16 \, \text{ft/sec}$. This means that when $t = 2$, $s = 101$, and $v(2) = 16$. **Q: How do we determine the initial height and velocity of the object?** ------------------------------------------------------------------- A: We can use the given information to determine the initial height and velocity of the object. Let's assume that the object's height at time $t = 0$ is $s(0)$. Then, we can write: $s(2) = s(0) + \int_{0}^{2} v(t) \, dt

where v(t)v(t) is the velocity of the object at time tt.

Q: How do we determine the velocity function?

A: We can use the fact that the object is moving up at a velocity of 16 ft/sec16 \, \text{ft/sec} after 2 seconds to determine the velocity function. Let's assume that the velocity function is a linear function of the form:

v(t)=mt+bv(t) = mt + b

where mm and bb are constants.

Q: How do we determine the acceleration function?

A: We can use the fact that the object is moving up at a velocity of 16 ft/sec16 \, \text{ft/sec} after 2 seconds to determine the acceleration function. Let's assume that the acceleration function is a constant function with value gg, where gg is the acceleration due to gravity.

Q: How do we determine the position function?

A: We can use the velocity and acceleration functions to determine the position function s(t)s(t). Let's assume that the position function is a quadratic function of the form:

s(t)=at2+bt+cs(t) = at^2 + bt + c

where aa, bb, and cc are constants.

Q: How do we fill in the blanks in the equation?

A: We can use the position function to fill in the blanks in the equation:

f(t)=expression in terms of tf(t) = \boxed{\text{expression in terms of } t}

Let's assume that the position function is:

s(t)=16t2+64t+81s(t) = 16t^2 + 64t + 81

Then, we can write:

f(t)=16t2+64t+81f(t) = 16t^2 + 64t + 81

Q: What is the final answer?

A: The final answer is:

f(t)=16t2+64t+81f(t) = 16t^2 + 64t + 81

This function describes the object's height above the ground at any given time tt.

Q: What is the significance of this problem?

A: This problem is significant because it demonstrates how to use given information to determine a function that describes a physical phenomenon. In this case, we used the given information to determine the function f(t)f(t) that describes the object's height above the ground at any given time tt.

Q: How can this problem be applied in real-life situations?

A: This problem can be applied in real-life situations where we need to determine the height of an object above the ground at any given time. For example, in physics, we may need to determine the height of a projectile above the ground at any given time. In engineering, we may need to determine the height of a building or a bridge above the ground at any given time.

Q: What are some common mistakes to avoid when solving this problem?

A: Some common mistakes to avoid when solving this problem include:

  • Not using the given information to determine the initial height and velocity of the object.
  • Not using the velocity and acceleration functions to determine the position function.
  • Not filling in the blanks in the equation correctly.
  • Not checking the units of the answer.

By avoiding these common mistakes, we can ensure that we arrive at the correct solution to the problem.