Putting It Together: Quadratic Functions
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- What is the maximum height achieved by the stone during its flight to the enemy castle?
- When does that maximum occur?
- How long would it take before the stone comes crashing down to Earth (in case it missed the castle altogether)?
[latex]h(t)=-{\large\frac{g}{2}}t^2+v_0t+h_0[/latex]
Here, [latex]h(t)[/latex] stands for the height at time [latex]t[/latex], [latex]v_0[/latex] is the initial upward velocity (speed), [latex]h_0[/latex] is the initial height of the projectile, and [latex]g[/latex] is a constant called the acceleration due to gravity. Near the surface of the Earth, [latex]g\approx9.8[/latex] meters per second squared, so [latex]-{\Large\frac{g}{2}}\approx-4.9[/latex]. Now putting the given values into their proper places, we have:[latex]h(t)=-4.9t^2+24.5t+8[/latex]
This is a quadratic function with [latex]a=-4.9[/latex], [latex]b=24.5[/latex], [latex]c=8[/latex]. The graph is a parabola, and it will have a maximum because [latex]a<0[/latex]. The maximum function value in this case represents the maximum height of the stone.Now we can answer the first two questions, What is the maximum height achieved by the stone during its flight to the enemy castle, and at what time does this occur? To find the maximum height and time at which it occurs, we use the vertex formula which will give [latex](t_\text{max},h(t_\text{max}))[/latex]. the time at maximum point and height at the time of the maximum point.
[latex]t_\text{max}=-{\large\frac{b}{2a}}=-{\large\frac{24.5}{2(-4.9)}}=2.5[/latex]
[latex]h(t_\text{max})=h\left(-{\large\frac{b}{2a}}\right)=h(2.5)=-4.9(2.5)^2+24.5(2.5)+8\approx38.6[/latex]
Therefore, the maximum height will be 38.6 meters, which occurs 2.5 seconds after launch. That’s a high-flying projectile!
Finally, to determine when the stone hits the ground, we just have to find the [latex]x[/latex]-intercept of the function. In other words, we have to solve [latex]h(t)=0[/latex]. That’s a job for the quadratic formula.
[latex]t_\text{final}={\large\frac{-b\pm\sqrt{b^2-4ac}}{2a}}={\large\frac{-24.5\pm\sqrt{24.5^2-4(4.9)(8)}}{2a}}\approx-0.3,5.3[/latex]
As expected, the quadratic formula gives us two answers, but only the positive one makes sense in this context. The stone lands on the ground about 5.3 seconds after it was launched.
Quadratic functions can be used to model the behavior of objects in free fall, amongst other things. We can use algebra to analyze this behavior for interesting features.
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- Putting It Together: Quadratic Functions. Authored by: Lumen Learning. License: CC BY: Attribution.
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- Trebuchet. Located at: https://upload.wikimedia.org/wikipedia/commons/2/2e/Trebuchet.jpg. License: Public Domain: No Known Copyright.