Math Problem Statement

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.

Solution

Let's break down the questions based on the given function y=f(t)=16t2+43t+37y = f(t) = -16t^2 + 43t + 37, where yy represents the height of the ball above the ground (in feet) at time tt 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=0t = 0. Substituting t=0t = 0 into the equation:

y(0)=16(0)2+43(0)+37=37 feety(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=0t = 0 and t=1t = 1 is given by the change in height divided by the change in time. We need to find the height at t=0t = 0 and t=1t = 1.

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

Now, the average velocity is:

Average velocity=y(1)y(0)10=64371=27 feet/second\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=1t = 1 second.

The instantaneous velocity is the derivative of the height function y=16t2+43t+37y = -16t^2 + 43t + 37. The derivative, v(t)v(t), is:

v(t)=ddt[16t2+43t+37]=32t+43v(t) = \frac{d}{dt} [-16t^2 + 43t + 37] = -32t + 43

Substitute t=1t = 1:

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

This confirms that the velocity at t=1t = 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 t2t^2 is negative). The time at which the maximum height occurs is given by:

tmax=b2a=432(16)=43321.34 secondst_{\text{max}} = \frac{-b}{2a} = \frac{-43}{2(-16)} = \frac{43}{32} \approx 1.34 \text{ seconds}

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

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

Let me calculate this.The maximum height of the ball occurs at approximately t=1.34t = 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 t2t^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.

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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