Question

a toy rocket is launched from the top of a building 139 feet tall at an initial velocity of 201 feet per second. a) give the function that describes the height of the rocket in terms of time t. b) determine the time at which the rocket reaches its maximum height, and the maximum height in feet. c) for what time interval will the rocket be more than 387 feet above ground level? d) after how many seconds will it hit the ground? a) the function that describes the height of the rocket in terms of t is s(t)=. b) the rocket reaches its maximum height of feet after approximately seconds. (round to the nearest hundredth as needed.) c) the rocket will be more than 387 feet above ground level for all values of t in the interval. (type your answer in interval notation. do not round until the final answer. then round to the nearest hundredth as needed.) d) the rocket will hit the ground after seconds. (round to the nearest hundredth as needed.)

Explanation:

Step1: Determine the height function

The general formula for the height of an object in vertical - motion is \(s(t)=-16t^{2}+v_{0}t + s_{0}\), where \(v_{0}\) is the initial velocity and \(s_{0}\) is the initial height.
Given \(v_{0} = 201\) and \(s_{0}=139\), the function is \(s(t)=-16t^{2}+201t + 139\).

Step2: Find the time of maximum height

For a quadratic function \(y = ax^{2}+bx + c\), the \(x\) - coordinate of the vertex (in our case, \(t\) for maximum height) is given by \(t=-\frac{b}{2a}\).
Here \(a=-16\) and \(b = 201\). So \(t=-\frac{201}{2\times(-16)}=\frac{201}{32}\approx6.28\) seconds.
Substitute \(t = \frac{201}{32}\) into \(s(t)\):

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Step3: Solve \(s(t)>387\)

Set \(-16t^{2}+201t + 139>387\).
\(-16t^{2}+201t + 139 - 387>0\), so \(-16t^{2}+201t - 248>0\).
Multiply both sides by \(- 1\) (and reverse the inequality sign): \(16t^{2}-201t + 248<0\).
Using the quadratic formula \(t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) for \(ax^{2}+bx + c = 0\) (here \(a = 16\), \(b=-201\), \(c = 248\)).
\(t=\frac{201\pm\sqrt{(-201)^{2}-4\times16\times248}}{2\times16}=\frac{201\pm\sqrt{40401-15872}}{32}=\frac{201\pm\sqrt{24529}}{32}=\frac{201\pm156.62}{32}\).
\(t_1=\frac{201 + 156.62}{32}\approx11.18\) and \(t_2=\frac{201-156.62}{32}\approx1.39\).
The solution of \(16t^{2}-201t + 248<0\) is \(t\in(1.39,11.18)\).

Step4: Solve \(s(t)=0\)

Set \(-16t^{2}+201t + 139 = 0\).
Using the quadratic formula \(t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) with \(a=-16\), \(b = 201\), \(c = 139\).
\(t=\frac{-201\pm\sqrt{201^{2}-4\times(-16)\times139}}{2\times(-16)}=\frac{-201\pm\sqrt{40401 + 8896}}{-32}=\frac{-201\pm\sqrt{49297}}{-32}\).
We take the positive root \(t=\frac{-201-\sqrt{49297}}{-32}\approx12.86\) (since \(t>0\)).

Answer:

a) \(s(t)=-16t^{2}+201t + 139\)
b) The rocket reaches its maximum height of \(770.27\) feet after approximately \(6.28\) seconds.
c) \((1.39,11.18)\)
d) \(12.86\)