The height of the football above the ground as a function of time \( h(t) \) is given by the equation
\[ h(t) = -16t^2 + 96t + 4 \]
where:
- \( h(t) \) is the height in feet,
- \( t \) is the time in seconds,
- the term \( -16t^2 \) represents the effect of gravity (in feet per second squared),
- \( 96t \) represents the initial upward velocity multiplied by time,
- \( 4 \) is the initial height from which the football is kicked.
Analysis of the Function
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Initial Height: At \( t = 0 \): \[ h(0) = -16(0)^2 + 96(0) + 4 = 4 \text{ feet} \] This confirms that the football is kicked from a height of 4 feet.
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Finding the Time of Maximum Height: To find when the football reaches its maximum height, we can use the vertex formula for a quadratic function \( h(t) = at^2 + bt + c \), where the maximum (vertex) occurs at \[ t = -\frac{b}{2a} \] Here, \( a = -16 \) and \( b = 96 \): \[ t = -\frac{96}{2(-16)} = \frac{96}{32} = 3 \text{ seconds} \]
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Maximum Height: We can substitute \( t = 3 \) back into the height equation to find the maximum height: \[ h(3) = -16(3)^2 + 96(3) + 4 \] \[ = -16(9) + 288 + 4 \] \[ = -144 + 288 + 4 = 148 \text{ feet} \]
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Finding When the Football Hits the Ground: To find when the football hits the ground, we set \( h(t) = 0 \): \[ -16t^2 + 96t + 4 = 0 \] We can use the quadratic formula \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Here, \( a = -16 \), \( b = 96 \), and \( c = 4 \): \[ t = \frac{-96 \pm \sqrt{96^2 - 4(-16)(4)}}{2(-16)} \] \[ t = \frac{-96 \pm \sqrt{9216 + 256}}{-32} \] \[ t = \frac{-96 \pm \sqrt{9472}}{-32} \] \[ \text{Calculating } \sqrt{9472} \approx 97.35 \] \[ t = \frac{-96 \pm 97.35}{-32} \] Considering the positive root: \[ t \approx \frac{1.35}{-32} \approx -0.042 \quad (\text{discard this, as time cannot be negative}) \]
For the negative root: \[ t \approx \frac{-193.35}{-32} \approx 6.05 \text{ seconds} \]
Summary
- The maximum height is 148 feet, reached at 3 seconds.
- The football hits the ground approximately 6.06 seconds after being kicked.
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