Solution
- The equation modeling the height of the falling object is:
[tex]h(t)=400-39t-157e^{-0.25t}[/tex]Question 1:
- To find the velocity of the falling object, we simply differentiate the function with respects to time (t).
- After this, we can find when the velocity of the object is -30ft/s by substituting V = -30 and find the value of t.
- Thus, we have:
[tex]\begin{gathered} h(t)=400-39t-157e^{-0.25t} \\ \\ h^{\prime}(t)=\frac{d}{dt}(h(t))=\frac{d}{dt}(400-39t-157e^{-0.25t}) \\ \\ h^{\prime}(t)=-39-157(-0.25)e^{-0.25t} \\ \\ h^{\prime}(t)=V(t)=-39+39.25e^{-0.25t} \\ \\ \text{ Now that we have the expression for }V(t),\text{ we can proceed to find when the velocity is -30 by making }V(t)=-30 \\ \\ -30=-39+39.25e^{-0.25t} \\ \text{ Add 39 to both sides} \\ 39.25e^{-0.25t}=39-30=9 \\ \\ \text{ Divide both sides by 39.25} \\ \\ e^{-0.25t}=\frac{9}{39.25} \\ \\ \text{ Take the natural log of both sides} \\ \\ -0.25t=\ln(\frac{9}{39.25}) \\ \\ \text{ Divide both sides by -0.25} \\ \\ t=-\frac{1}{0.25}\ln(\frac{9}{39.25}) \\ \\ t=5.890907...\approx5.89s \end{gathered}[/tex]- The time is 5.89s
Question 2:
- The height of the object at time t = 5.89s is gotten by substituting this value of time (t) into the equation of the height given to us.
- We have:
[tex]\begin{gathered} h(t)=400-39t-157e^{-0.25t} \\ \\ put\text{ }t=5.89 \\ h(t)=400-39(5.89)-157e^{-0.25(5.89)} \\ \\ h(t)=134.2818...\approx134ft\text{ }(\text{ To the nearest foot\rparen} \end{gathered}[/tex]- the height is 134 ft