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Sagot :
Given the equation:
[tex]h(t)=-4.9t^2+10t+0.1[/tex]Where h represents the height of water above the basin and t tracks time in seconds.
Let's answer the following questions:
• Part 1.
Find the maximum height h of the water above the basin.
To find the maximum height, take the derivative of h(t):
[tex]\begin{gathered} h^{\prime}(t)=\frac{d}{dt}(-49t^2+10t+0.1) \\ \\ h^{\prime}(t)=98t+10 \\ \\ v(t)=h^{\prime}(t)=98t+10 \end{gathered}[/tex]Where v(t) represents the velocity.
At maximum height, the velcoity is zero.
Now, plug in 0 for v(t) and solve for the time:
[tex]\begin{gathered} 0=-98t+10 \\ \\ 98t=10 \\ \\ t=\frac{10}{98} \\ \\ t=\frac{5}{49} \end{gathered}[/tex]Now plug in 5/49 for t in h(t) and solve:
[tex]\begin{gathered} h(\frac{5}{49})=-49(\frac{5}{49})^2+10(\frac{5}{49})+0.1 \\ \\ h(\frac{5}{49})=-\frac{25}{49}+\frac{50}{49}+0.1 \\ \\ h(\frac{5}{49})=0.610 \end{gathered}[/tex]Therefore, the maximum height of the water above the basin is 0.61 feet.
• Part 2.
Find the time t it takes for water that exits the spout to land in the basin.
When the water lands the basin, the height, h(t) = 0.
To find the time, plug in 0 for h(t) and solve for t.
We have:
[tex]\begin{gathered} 0=-49t^2+10t+0.1 \\ \\ 49t^2-10t-0.1=0 \end{gathered}[/tex]SOlve uisng the quadratic formula:
Where:
a = 49
b = -10
c = -0.1
We have:
[tex]\begin{gathered} t=\frac{-b\pm\sqrt{b^2-4ac}}{2a} \\ \\ t=\frac{-49\pm\sqrt{-10^2-4(49)(-0.1)}}{2(49)} \\ \\ t=\frac{-49\pm\sqrt{100+19.6}}{98} \\ \\ t=\frac{-49\pm\sqrt{119.6}}{98} \\ \\ t=\frac{-49\pm10.94}{98} \\ \\ t=\frac{-49+10.94}{98},\frac{-49-10.94}{98} \\ \\ t=0.214,\text{ -0.0096} \end{gathered}[/tex]Since the time cannot be negative, let's take the positive value.
Therefore, the time is 0.214 seconds.
ANSWER:
Part 1: 0.610 feet
Part 2: 0.214 seconds
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