Any object influenced by gravity falls freely according to the equation: h(t) = -16t² + v₀t + h₀ , where v₀ is initial velocity, and h₀ is the initial height above the ground.

Our ball's motion can be described by the quadratic function h(t) = -16t² + 57.3t + 4, since our initial velocity is 57.3 ft/s (from the previous section). The initial height of the ball thrown in the previous section is 4 ft above the ground.

Now, our focus for this resource is the solutions to this quadratic when h(t) = 0, and more importantly, the meaning. Use your graphing calculator to find the roots of this equation.

How do we find roots?

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Solve 0 = -16t² + 57.3t + 4.
Or, find where the graph h(t) = -16t² + 57.3t + 4 crosses the horizontal axis.

If you don't have a graphing calculator handy, click here to access a website that will generate roots for your quadratic function.

Once you have found the roots,

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Check to see if you are correct.

(-0.07,0) and (3.65,0)

Since this is a vertical height function, what does the root (3.65,0) tell us?

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The root (3.65,0) tells us the ball is in the air for 3.65 seconds before it reaches the ground (if it isn't caught, but travels all the way to the ground).

True or false: The negative solution is valid for this situation?

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False! The negative root (-0.07,0) is an extraneous root that doesn't really tell us anything useful for this example.

If we launch at an initial time of t₀ = 0, how long is the ball in the air?

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3.65 seconds

Although it is important to be able to solve mathematical equations, the real significance lies in interpreting the results in a realistic setting such as this.