A projectile is fired vertically with an initial velocity of 576 feet per second. The distance in feet above the ground after seconds is given by the equation . For what
range of time (measured in seconds) will the height be greater than 5040 feet?

Ask by Bright Chadwick. in the United States

Dec 16,2024

Question

A projectile is fired vertically with an initial velocity of 576 feet per second. The distance in feet above the ground after seconds is given by the equation . For what
range of time (measured in seconds) will the height be greater than 5040 feet?

Ask by Bright Chadwick. in the United States

Dec 16,2024

UpStudy AI · Accepted Answer

To find the range of time for which the height $$ s $$ will be greater than 5040 feet, we need to solve the inequality $$ s > 5040 $$ using the given equation $$ s = -16t^2 + 576t $$. Given: - Initial velocity: 576 feet per second - Equation for distance above the ground: $$ s = -16t^2 + 576t $$ - Height $$ s $$ should be greater than 5040 feet We need to solve the inequality $$ -16t^2 + 576t > 5040 $$ to find the range of time $$ t $$. Let's solve this inequality to find the range of time. Solve the inequality by following steps: - step0: Solve the inequality by testing the values in the interval: $$-16t^{2}+576t>5040$$ - step1: Move the expression to the left side: $$-16t^{2}+576t-5040>0$$ - step2: Rewrite the expression: $$-16t^{2}+576t-5040=0$$ - step3: Factor the expression: $$-16\left(t-21 ight)\left(t-15 ight)=0$$ - step4: Divide the terms: $$\left(t-21 ight)\left(t-15 ight)=0$$ - step5: Separate into possible cases: $$\begin{align}&t-21=0\&t-15=0\end{align}$$ - step6: Solve the equation: $$\begin{align}&t=21\&t=15\end{align}$$ - step7: Determine the test intervals: $$\begin{align}&t<15\&1521\end{align}$$ - step8: Choose a value: $$\begin{align}&t_{1}=14\&t_{2}=18\&t_{3}=22\end{align}$$ - step9: Test the chosen value: $$\begin{align}&t<15 extrm{ } extrm{is not a solution}\&1521 extrm{ } extrm{is not a solution}\end{align}$$ - step10: The final solution is $$15 5040 $$ is $$ 15 < t < 21 $$. Therefore, the height $$ s $$ will be greater than 5040 feet for a range of time from 15 seconds to 21 seconds. So, the range of time for which the height $$ s $$ will be greater than 5040 feet is $$ 15 < t < 21 $$. The range of time is $$ 15 < t < 21 $$ seconds.

A projectile is fired vertically with an initial velocity of 576 feet per second. The distance in feet above the ground after seconds is given by the equation . For what
range of time (measured in seconds) will the height be greater than 5040 feet?

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A projectile is fired vertically with an initial velocity of 576 feet per second. The distance in feet above the ground after seconds is given by the equation . For what range of time (measured in seconds) will the height be greater than 5040 feet?

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A projectile is fired vertically with an initial velocity of 576 feet per second. The distance in feet above the ground after seconds is given by the equation . For what
range of time (measured in seconds) will the height be greater than 5040 feet?