Question

5.1: homework assignment
score: 23.6/29 answered: 24/29
question 19
an object is thrown upward at a speed of 170 feet per second by a machine from a height of 5 feet off the ground. the height h of the object after t seconds can be found using the equation h = -16t² + 170t + 5
when will the height be 289 feet?
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when will the object reach the ground?
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Explanation:

Step1: Set up equation for height of 289 feet

Set $h = 289$ in $h=-16t^{2}+170t + 5$, so $-16t^{2}+170t + 5=289$. Rearrange to get $-16t^{2}+170t-284 = 0$. Divide by -2: $8t^{2}-85t + 142=0$.

Step2: Use quadratic formula

For a quadratic equation $ax^{2}+bx + c = 0$ ($a = 8$, $b=-85$, $c = 142$), the quadratic formula is $t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. Calculate $b^{2}-4ac=(-85)^{2}-4\times8\times142=7225 - 4544=2681$. Then $t=\frac{85\pm\sqrt{2681}}{16}$. $\sqrt{2681}\approx51.78$, so $t_1=\frac{85 + 51.78}{16}\approx8.55$ and $t_2=\frac{85 - 51.78}{16}\approx2.08$.

Step3: Set up equation for reaching ground

When the object reaches the ground, $h = 0$. So $-16t^{2}+170t + 5=0$. Here $a=-16$, $b = 170$, $c = 5$.

Step4: Use quadratic formula again

$b^{2}-4ac=(170)^{2}-4\times(-16)\times5=28900 + 320=29220$. $t=\frac{-170\pm\sqrt{29220}}{-32}$. $\sqrt{29220}\approx170.94$. $t=\frac{-170\pm170.94}{-32}$. We take the positive root $t=\frac{-170 + 170.94}{-32}$ is negative, and $t=\frac{-170-170.94}{-32}=\frac{340.94}{32}\approx10.65$.

Answer:

When the height is 289 feet, $t\approx2.08$ seconds and $t\approx8.55$ seconds. When the object reaches the ground, $t\approx10.65$ seconds.