If the object in Problem 2 is released from rest 30ftabove the ground instead of 500ft, when will it strike the ground? [ Hint:Use Newton’s method to solve for t ]

Use Newton’s method to solve for t.

$t_{n+1}=t_n-\frac{f\left(t_n\right)}{f\text{'}\left(t_n\right)}$

For finding the weight of the object apply:

$\text{Net} \text{force}=W-\text{Drag} \text{force}$

role="math" localid="1663944542020" $$\begin{gathered} ma=W-10v \\ m\frac{dv}{dt}=4W-10v \\ W=mg\left(a=\frac{dv}{dt}\right) \\ 400=32m \end{gathered}$$

Apply formula for finding the velocity:

$$\begin{gathered} m\frac{dv}{dt}=400-10v \\ 12.5\frac{dv}{dt}=400-10v \\ \frac{dv}{dt}=32-0.8v \\ v.e^{0.8t}=\int 32e^{0.8t}dt \left(\text{Integrating factor} e^{0.8t}\right) \end{gathered}$$

Further, solve the above expression

role="math" localid="1663944724324" $$\begin{gathered} v.e^{0.8t}=40e^{0.8t}+C \left(c=\text{integration constant}\right) \\ v=40+C{e}^{-0.8t} \\ Atv=0,t=0, then C=-40 \\ v=40-40e^{-0.8t} \end{gathered}$$

$$\begin{gathered} v=\frac{dx}{dt} \\ \frac{dx}{dt}=40-40e^{-0.8t} \\ x\left(t\right)=40t+50e^{-0.8t}+C \end{gathered}$$

Put the value of$t=0,x=0, then C=-20$

$x\left(t\right)=40t+50e^{-0.8t}-20$

$x\left(t\right)=40t+50e^{-0.8t}-20$

When object hits the ground $x=0$.

$40t+50e^{-0.8t}=20$

By solving trial and error method$t=1.67\sec .$

Therefore, the value of t is$t=1.67\sec .$