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Let's break down the questions based on the given function \( y = f(t) = -16t^2 + 43t + 37 \), where \( y \) represents the height of the ball above the ground (in feet) at time \( t \) 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 \( t = 0 \). Substituting \( t = 0 \) into the equation:

\[
y(0) = -16(0)^2 + 43(0) + 37 = 37 \text{ feet}
\]

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 \( t = 0 \) and \( t = 1 \) is given by the change in height divided by the change in time. We need to find the height at \( t = 0 \) and \( t = 1 \).

\[
y(0) = 37
\]
\[
y(1) = -16(1)^2 + 43(1) + 37 = -16 + 43 + 37 = 64
\]

Now, the average velocity is:

\[
\text{Average velocity} = \frac{y(1) - y(0)}{1 - 0} = \frac{64 - 37}{1} = 27 \text{ feet/second}
\]

This confirms that the average velocity is **27 feet/second**.

### Part C: Approximate the velocity of the ball at \( t = 1 \) second.
The instantaneous velocity is the derivative of the height function \( y = -16t^2 + 43t + 37 \). The derivative, \( v(t) \), is:

\[
v(t) = \frac{d}{dt} [-16t^2 + 43t + 37] = -32t + 43
\]

Substitute \( t = 1 \):

\[
v(1) = -32(1) + 43 = -32 + 43 = 11 \text{ feet/second}
\]

This confirms that the velocity at \( t = 1 \) 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 \( t^2 \) is negative). The time at which the maximum height occurs is given by:

\[
t_{\text{max}} = \frac{-b}{2a} = \frac{-43}{2(-16)} = \frac{43}{32} \approx 1.34 \text{ seconds}
\]

Now, substitute this \( t_{\text{max}} \) back into the height function to find the maximum height:

\[
y(1.34) = -16(1.34)^2 + 43(1.34) + 37
\]

Let me calculate this.The maximum height of the ball occurs at approximately \( t = 1.34 \) 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:

1. How would you determine the time at which the ball hits the ground?
2. What is the velocity of the ball at its maximum height?
3. How long does the ball stay in the air before hitting the ground?
4. How would you interpret the meaning of the derivative of the height function physically?
5. How would the motion of the ball change if the coefficient of \( t^2 \) 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.

A ball is shot into the air from a bridge, and its height, y, in feet, above the ground t seconds after it is thrown is given by y = f(t) = -16t^2 + 43t + 37. Answer the following questions. A. How high above the ground is the bridge? B. What is the average velocity of the ball during the first second? C. Approximate the velocity of the ball at t = 1 second. D. Graph the function f(t) in your calculator. From the graph estimate the maximum height the ball will reach.

Solve a quadratic function to find the height of a ball shot from a bridge, determine its average and instantaneous velocities, and graph its maximum height. Explore key algebraic concepts like derivatives, average velocity, and the quadratic vertex formula. Suitable for Grades 10-12.

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