Using A Quadratic Equation To Solve Ball Thrown Vertically ✓ Solved

Using a Quadratic Equation Problems A ball is thrown vertically upward from the top of a

A ball is thrown vertically upward from the top of a building 112 feet tall with an initial velocity of 96 feet per second. The distance s (in feet) of the ball from the ground after t seconds is given by the quadratic equation: s(t) = 112 + 96t - 16t². Complete the table of values for s(t) and interpret each point. Additionally, answer the following questions:

  1. After how many seconds does the ball strike the ground?
  2. After how many seconds will the ball pass the top of the building on its way down?
  3. How long will it take the ball to reach the maximum height?
  4. What is the maximum height?

Sample Paper For Above instruction

The motion of an object thrown vertically upward can be effectively modeled using a quadratic equation due to the constant acceleration of gravity. In this case, we analyze a ball thrown from a height of 112 feet with an initial upward velocity of 96 ft/sec. Its height relative to the ground after t seconds is described by the equation s(t) = 112 + 96t - 16t². This quadratic function encapsulates the initial position, initial velocity, and the constant acceleration due to gravity (approximated to 32 ft/sec²), which influences the downward pull of gravity on the ball's trajectory.

Completing the table and interpreting each point

To analyze the motion, it is extensive to compute s(t) for different values of t. Let us select key time points—such as when the ball reaches the ground, the maximum height, and intermediate times—to gather data points.

At t = 0 seconds:

s(0) = 112 + 96(0) - 16(0)^2 = 112 feet.

This signifies the initial position of the ball, which is at the top of the building.

At t = 3 seconds:

s(3) = 112 + 96(3) - 16(3)^2 = 112 + 288 - 144 = 256 feet.

The ball has moved upward, reaching a height greater than the initial height.

At t = 6 seconds:

s(6) = 112 + 96(6) - 16(6)^2 = 112 + 576 - 576 = 112 feet.

The ball descends back to the height of the building's top.

At t = 7 seconds:

s(7) = 112 + 96(7) - 16(7)^2 = 112 + 672 - 784 = 0 feet.

This indicates the ball has hit the ground.

Testing t = 0.5 seconds:

s(0.5) = 112 + 96(0.5) - 16(0.5)^2 = 112 + 48 - 4 = 156 feet.

The ball is ascending, moving higher than the initial height.

Each data point signifies the ball's height at specific times, with the maximum height occurring at the vertex of the parabola described by s(t).

The interpretation of each point corresponds to the physical position of the ball at those times, which elucidates the upward motion, the height reached, and the descent back to the ground.

Solving the specific questions

1. When does the ball strike the ground?

To determine when the ball hits the ground, set s(t) = 0 and solve:

0 = 112 + 96t - 16t².

Multiplying both sides by -1 to simplify:

0 = -112 - 96t + 16t².

Rewrite as a standard quadratic form:

16t² - 96t - 112 = 0.

Divide throughout by 16:

t² - 6t - 7 = 0.

Apply the quadratic formula:

t = [6 ± √(36 + 28)] / 2

t = [6 ± √64] / 2

t = [6 ± 8] / 2.

Two solutions:

t = (6 + 8)/2 = 14/2 = 7 seconds,

t = (6 - 8)/2 = -2/2 = -1 seconds (discard as time cannot be negative).

Therefore, the ball hits the ground after 7 seconds.

2. When does the ball pass the top of the building on the way down?

The ball passes the top of the building when s(t) = 112 feet, corresponding to the initial height. To find when it occurs after reaching peak height, solve:

112 + 96t - 16t² = 112.

Subtract 112 from both sides:

96t - 16t² = 0.

Factor:

16t(6 - t) = 0.

Solutions:

t = 0 (initial time),

t = 6 seconds.

Time t = 0 corresponds to the starting point; the other solution indicates the ball passes the top again on its way down after 6 seconds.

3. How long does it take the ball to reach maximum height?

The maximum height corresponds to the vertex of the parabola s(t). For a quadratic equation in the form s(t) = at² + bt + c, the vertex occurs at t = -b/(2a).

In our equation:

a = -16,

b = 96.

So,

t = -96 / (2 * -16) = -96 / -32 = 3 seconds.

Thus, the ball reaches its maximum height after 3 seconds.

4. What is the maximum height?

To find the maximum height, substitute t = 3 into s(t):

s(3) = 112 + 96(3) - 16(3)^2 = 112 + 288 - 144 = 256 feet.

The maximum height achieved by the ball is 256 feet.

Conclusion

This analysis demonstrates how quadratic equations effectively model the motion of objects under constant acceleration, such as the gravitational pull on a projectile. Understanding the vertex and roots of these quadratics allows precise determination of key points in motion—like maximum height, time to reach it, and total time of flight. Recognizing these features is essential in physics and engineering applications where calculating projectile trajectories is necessary.

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