Question

question 16
a rocket is launched, and its height above sea level t seconds after launch is given by the equation h(t)= - 4.9t² + 1500t + 300.
a) from what height was the rocket launched?
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b) what is the maximum height the rocket reaches?
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c) if the rocket will splash down in the ocean, when will it splash down?
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question 17
a person standing close to the edge on top of a 96 - foot building throws a ball vertically upward. the quadratic function h(t)= - 16t² + 116t + 96 models the balls height about the ground, h(t), in feet, t seconds after it was thrown.
a) what is the maximum height of the ball? (round your answer to the nearest hundredths.)
feet
b) how many seconds does it take until the ball hits the ground?
seconds
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question 18

Explanation:

Step1: Find the launch - height of the rocket

The rocket's height function is $h(t)=-4.9t^{2}+1500t + 300$. At the time of launch, $t = 0$. Substitute $t = 0$ into the function: $h(0)=-4.9(0)^{2}+1500(0)+300$.

Step2: Find the maximum height of the rocket

For a quadratic function $y = ax^{2}+bx + c$ ($a
eq0$), the $t$ - value of the vertex (which gives the maximum or minimum) is $t=-\frac{b}{2a}$. Here, $a=-4.9$, $b = 1500$, $c = 300$. First, find $t=-\frac{1500}{2\times(-4.9)}=\frac{1500}{9.8}$. Then substitute this $t$ - value into $h(t)$ to get the maximum height: $h(\frac{1500}{9.8})=-4.9(\frac{1500}{9.8})^{2}+1500(\frac{1500}{9.8})+300$.

Step3: Find the splash - down time of the rocket

When the rocket splashes down, $h(t)=0$. So we need to solve the quadratic equation $-4.9t^{2}+1500t + 300 = 0$ for $t$. Use the quadratic formula $t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$, where $a=-4.9$, $b = 1500$, $c = 300$.

Step4: For the ball - find the maximum height

The ball's height function is $h(t)=-16t^{2}+116t + 96$, with $a=-16$, $b = 116$, $c = 96$. The $t$ - value of the vertex is $t=-\frac{b}{2a}=-\frac{116}{2\times(-16)}=\frac{116}{32}=\frac{29}{8}$. Substitute $t=\frac{29}{8}$ into $h(t)$: $h(\frac{29}{8})=-16(\frac{29}{8})^{2}+116(\frac{29}{8})+96$.

Step5: For the ball - find the time it hits the ground

When the ball hits the ground, $h(t)=0$. So we solve the quadratic equation $-16t^{2}+116t + 96 = 0$. Divide through by - 4 to simplify: $4t^{2}-29t - 24 = 0$. Then use the quadratic formula $t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$, where $a = 4$, $b=-29$, $c=-24$.

Answer:

Question 16:
a) The rocket was launched from a height of 300 feet.
b) First, $t=\frac{1500}{9.8}\approx153.06$. Then $h(\frac{1500}{9.8})=-4.9(\frac{1500}{9.8})^{2}+1500(\frac{1500}{9.8})+300\approx115122.45$ feet.
c) Using the quadratic formula for $-4.9t^{2}+1500t + 300 = 0$, $t=\frac{-1500\pm\sqrt{1500^{2}-4\times(-4.9)\times300}}{2\times(-4.9)}$. We take the positive root since time cannot be negative, $t\approx306.12$ seconds.
Question 17:
a) $t=\frac{29}{8}=3.625$. $h(3.625)=-16(3.625)^{2}+116(3.625)+96=-16\times13.140625+419.5 + 96=-210.25+419.5+96=209.25 + 96=305.25$ feet.
b) For $4t^{2}-29t - 24 = 0$, $t=\frac{29\pm\sqrt{(-29)^{2}-4\times4\times(-24)}}{2\times4}=\frac{29\pm\sqrt{841 + 384}}{8}=\frac{29\pm\sqrt{1225}}{8}=\frac{29\pm35}{8}$. We take the positive root $t=\frac{29 + 35}{8}=\frac{64}{8}=8$ seconds.