Question
Explanation
we are given the height function as:
[tex]h(t)=16.1t^2-1.75t^3\text{ }[/tex]
The first step will be to get the velocity function
[tex]h^{\prime}(t)=32.2t-5.25t^2[/tex]
The acceleration is given by
[tex]acceleration=h^{\prime}^{\prime}(t)=32.2-10.5t[/tex]
Since we are told to find how high the rocket will be at an instant of 10ft/s²
Therefore, we will equate the acceleration formula to 10ft/s² so that we will obtain t
[tex]\begin{gathered} 10=32.2-10.5t \\ 10.5t=32.2-10 \\ 10.5t=22.2 \\ t=2.11428 \end{gathered}[/tex]
The final step will be to substitute t =2.11428 into the initial height function
[tex]\begin{gathered} h=16.1\cdot\:2.11428^2-1.75\cdot\:2.11428^3=55.43027 \\ h=55.43\text{ ft} \\ \end{gathered}[/tex]
The answer is h = 55.43 ft
Part F
was when the rocket is moving 35ft/s, we will have to equate 35ft/s into the velocity formula
[tex]\begin{gathered} h^{\prime}(t)=32.2t-5.25t^2=35 \\ \\ t_{1,\:2}=\frac{-3220\pm \:140\sqrt{154}}{2\left(-525\right)} \\ \\ t=1.4120\:t=4.7212 \\ \\ \end{gathered}[/tex]
But we are told that t cannot be greater than 4
So that t=4.7212s is invalid
Thus, our value for t= 1.4120
So we will substitute this to get the acceleration
[tex]\begin{gathered} =32.2-10.5(1.4120) \\ =17.374ft\text{/s}^2 \end{gathered}[/tex]
The answer is 17.37 ft/s²
