Accounting for Air Resistance

Relevant wiki: Introduction to Kinematics

For situations with velocity-dependent forces, it is useful to employ the following strategy:

1) Start with Newton's Second law
2) Apply the differentiation - Chain rule
3) Separate variables and integrate

${F = m \frac{dv}{dt} = m \frac{dv}{dy} \frac{dy}{dt} = mv \frac{dv}{dy}}.$ $\begin{aligned} mv \frac{dv}{dy} &= -mg - \frac{v^2}{100} = -mg - \alpha v^{2}. \\ \text{Here}, \, \alpha \,&=\, \frac{1}{100}. \\ dy &= \frac{mv \, {dv}}{-mg - \alpha v^{2}}. \\ y_{max} &= \int_{v_0}^{0} \frac{mv \, {dv}}{-mg - \alpha v^{2}}. \\ &= \int_0^{v_0} \frac{mv \, {dv}}{mg + \alpha v^{2}}. \\ &=\frac{m}{2 \alpha} \int_{mg}^{mg + \alpha {v_0}^2} \frac{du}{u}. \\ &= \frac{m}{2 \alpha} \ln \left(\frac{mg + \alpha {v_0}^2}{mg}\right). \\ \end{aligned}$
Plugging in numbers gives $y_{max} = \boxed{120 m}$ , to the nearest meter.

We can define force as, $F=ma$ $F = -mg - \frac{v^2}{100}$ $a = -g - \frac{v^2}{m100}$ $a=v\frac{dv}{dx}$ At highest point $v=0$ , $\frac{vdv}{dx}=-\frac{\left(100g+v^2\right)}{100}$ $\frac{v}{\left(100g+v^2\right)}dv=-\frac{dx}{100}$ By integrating both sides we get, $\int _{100}^0\frac{v}{\left(1000+v^2\right)}dv=-\frac{1}{100}\int _0^xdx$ $\dfrac{\ln\left(1000\right)}{2}-\dfrac{\ln\left(11000\right)}{2}=-\frac{x}{100}$ $-\frac{x}{100}=-1.1989$ $x=119.89$ So, the answer is approximately $x=120 m$ .

I originally did the problem analytically, but since there are a lot of analytic solutions already, here's a solution that's done by computer simulation. I used Runge-Kutta 4th order to solve the second order ODE with time-step $10^{-6}$ :

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#4th order Runge-Kutta numerical solution of second order ODE

import math

#initialisation

time = 0
deltaT = 10**-6


u_1 = 0
u_2 = 100

def differentialEquation(t,x,xdot):  #applying RK4 by splitting 2nd order ODE into 2 first order ODEs
    return (-10 - ((xdot**2)/100))


while u_2 >= 0:

    #evaluating slope at various points


    k0 = deltaT*u_2
    L0 = deltaT*differentialEquation(time,u_1,u_2)

    k1 = deltaT*(u_2 + 0.5*L0)
    L1 = deltaT*differentialEquation(time + 0.5*deltaT, u_1 + 0.5*k0, u_2 + 0.5*L0)

    k2 = deltaT*(u_2 + 0.5*L1)
    L2 = deltaT*differentialEquation(time + 0.5*deltaT, u_1 + 0.5*k1, u_2 + 0.5*L1)

    k3 = deltaT*(u_2 + L2)
    L3 = deltaT*differentialEquation(time + deltaT, u_1 + k2, u_2 + L2)


    #RK4 approximation of values

    u_1 = u_1 + (1/6)*(k0 + 2*k1 + 2*k2 + k3)
    u_2 = u_2 + (1/6)*(L0 + 2*L1 + 2*L2 + L3)

    time += deltaT



print("maximum height: " + str(u_1))

For variety, the differential equation $\dot{v} \; = \; -10 - \frac{v^2}{100} \hspace{2cm} v(0) = 100$ can be solved to obtain $v \; = \; 10\sqrt{10}\tan\big(\alpha - \tfrac{t}{\sqrt{10}}\big) \hspace{2cm} 0 \le t \le \alpha\sqrt{10}$ where $\tan\alpha = \sqrt{10}$ . The the height of the particle at time $t$ is $\int_0^t 10\sqrt{10}\tan\big(\alpha - \tfrac{u}{\sqrt{10}}\big)\,du \; = \; 100\ln\left(\frac{\sec\alpha}{\sec\big(\alpha - \tfrac{t}{\sqrt{10}}\big)}\right) \hspace{1cm} 0 \le t \le \alpha\sqrt{10}$ The particle reaches maximum height $100\sec\alpha = 50\ln11 \approx \boxed{120}$ when $t = \alpha\sqrt{10}$ .