Math Problem Statement
Solution
Let's break down the questions based on the given function , where represents the height of the ball above the ground (in feet) at time seconds.
Part A: How high above the ground is the bridge?
The height of the bridge is given by the constant term in the function when . Substituting into the equation:
This confirms the height of the bridge is 37 feet.
Part B: What is the average velocity of the ball during the first second?
The average velocity between and is given by the change in height divided by the change in time. We need to find the height at and .
Now, the average velocity is:
This confirms that the average velocity is 27 feet/second.
Part C: Approximate the velocity of the ball at second.
The instantaneous velocity is the derivative of the height function . The derivative, , is:
Substitute :
This confirms that the velocity at second is 11 feet/second.
Part D: Graph the function and find the maximum height.
To find the maximum height, we can calculate the vertex of the parabola, since the function is quadratic and opens downward (because the coefficient of is negative). The time at which the maximum height occurs is given by:
Now, substitute this back into the height function to find the maximum height:
Let me calculate this.The maximum height of the ball occurs at approximately seconds, and the maximum height reached is approximately 65.9 feet.
Would you like further details or clarification on any of these parts?
Here are 5 additional related questions for deeper understanding:
- How would you determine the time at which the ball hits the ground?
- What is the velocity of the ball at its maximum height?
- How long does the ball stay in the air before hitting the ground?
- How would you interpret the meaning of the derivative of the height function physically?
- How would the motion of the ball change if the coefficient of was smaller?
Tip: When analyzing quadratic functions like this, finding the vertex gives key information, such as the maximum or minimum value of the function.
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Math Problem Analysis
Mathematical Concepts
Quadratic Functions
Instantaneous Velocity
Average Velocity
Derivatives
Maximum Height
Formulas
y = -16t^2 + 43t + 37 (Position function for height)
v(t) = dy/dt = -32t + 43 (Instantaneous velocity)
Average velocity = (y(1) - y(0)) / (1 - 0)
Theorems
Quadratic Vertex Formula
Suitable Grade Level
Grades 10-12
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