Respuesta :
Answer:
• (a)112 feet
,• (b)1.28125 seconds.
,• (c)138.265625 feet.
,• (d)4.22091 seconds
Explanation:
The height of the rocket in terms of the number of seconds t since the rocket's engine stops firing is given below.
[tex]S\mleft(t\mright)=-16t^2+41t+112[/tex]Part A
At the time the rocket stopped firing, t=0.
[tex]S(0)=-16(0)^2+41(0)+112=112[/tex]The rocket was 112 feet above the ground when it stopped firing.
Part B
The value of t at which the rocket reaches its maximum height is the equation of the line of symmetry.
To find this equation, we use the formula below.
[tex]t=-\frac{b}{2a}=-\frac{41}{-2\times16}=1.28125\text{ seconds}[/tex]The rocket reaches its maximum height after 1.28125 seconds.
Part C
To find the maximum height, substitute t=1.28125 into S(t).
[tex]\begin{gathered} S\mleft(t\mright)=-16t^2+41t+112 \\ \implies S(1.28125)=-16(1.28125)^2+41(1.28125)+112 \\ =138.265625\text{ ft} \end{gathered}[/tex]The maximum height of the rocket is 138.265625 feet.
Part D
When the rocket hits the ground, the height is 0.
Set S(t)=0 and solve for t as follows.
[tex]S(t)=-16t^2+41t+112=0[/tex]Using the quadratic formula:
[tex]\begin{gathered} t=\dfrac{-41\pm\sqrt[]{41^2-4(-16)(112)}}{2\times-16}=\dfrac{-41\pm\sqrt[]{1681-(-7168)}}{-32} \\ =\dfrac{-41\pm\sqrt[]{1681+7168}}{-32} \\ =\dfrac{-41\pm\sqrt[]{8849}}{-32} \\ t=\dfrac{-41+\sqrt[]{8849}}{-32}\text{ or }t=\dfrac{-41-\sqrt[]{8849}}{-32} \\ t=-1.658\; \text{or }t=4.22091 \end{gathered}[/tex]Since t cannot be negative, the rocket will hit the ground after 4.22091 seconds.
