Is physics a drag?

The force on the ball are the one due to gravity and the other due to drag. As the ball is moving in upward direction, both these forces are acting in downward direction. Therefore,

$m \times \frac{dv}{dt}$ = $-mg - βv$

$\frac{dv}{dt} = -(\frac{βv}{m} + g)$

$\frac{dv}{dx} \times \frac{dx}{dt} = -(\frac{βv}{m} + g)$

$v \times \frac{dv}{dx} = -(\frac{βv}{m} + g)$

$\frac{v \times dv}{βv + mg} = -\frac{dx}{m}$

$\frac{dv}{β} - \frac{mg \times dv}{β(βv + mg)} = -\frac{dx}{m}$

Integrating both the sides, the LHS from 10 to 0, and the RHS from 0 to "h(max)"

$\frac{v}{β}$ (10 to 0) - $\frac{mg}{β} \times \frac{ln(βv + mg)}{β}$ (10 to 0) = $-\frac{x}{m}$ (o to h)

Substitution value of initial velocity, v = 10 m/s,

$\frac{10}{β} - \frac{mg}{β^{2}} \times {ln(10β + mg) - ln(mg)} = \frac{h}{m}$

Substituting other values,

$h = 10 - (9.8) \times ln(\frac{1 + 0.98}{0.98})$

$h = \boxed{3.108}$

This problem can be solved with the use of elementary, ordinary linear differential equations. We first have (saying that upwards is specified by $x>0$ ): $F_{net} = - F_g - F_\text{drag}$ Note that we have: $F_g=mg$ And $F_\text{drag}=\beta v=\beta \dot{x}$ Giving us an ODE: $m\ddot{x}=-mg-\beta\dot{x}$ Dividing through by $m$ and rearranging: $\ddot{x}+\frac{\beta}{m}\dot{x}+g=0$ Note that $\beta=m=0.1$ and that $g=9.8$ (for our sign convention), which gives us our final form to solve: $\ddot{x}+\dot{x}+9.8=0$ Using the standard ODE techniques, we let $x=e^{\alpha t}$ for which we solve: $\alpha^2+\alpha=0$ Gives us: $\alpha(\alpha+1)=0$ And: $\alpha=-1, 0$ A linear combination gives us: $c_1e^{-t}+c_2=x$ Clearly, the particular solution is then: $x_p=-9.807t$ Summing these two solutions: $c_1e^{-t}+c_2-9.807t=x(t)$ We are restricted by: $x(0)=0$ And: $\dot{x}(0)=10$ Finding the constants under these initial conditions gives us: $-19.8e^{-t}-9.8t+19.8=x(t)$ It should be clear, then, that the maximum is at $t=\log\left(\frac{99}{49}\right)$ , which is, once plugged in, approximately: $\max_{t\in[0, \infty)} x(t)\approx 3.1077$

$$ From the Newton's second law of motion , we obtain $$ $\begin{aligned} \sum F&=ma\\ -mg-\beta v&=m\frac{dv}{dt}\\ -mg-\beta v&=m\frac{dv}{dt}\cdot\frac{dy}{dy}\\ dy&=-\frac{mv}{mg+\beta v}dv\\ \int_{h_i}^{h_f} dy&=-\int_{v_i}^{v_f}\frac{mv}{mg+\beta v}dv\\ h_f-h_i&=\int_{v_f}^{v_i}\frac{mv}{mg+\beta v}dv. \end{aligned}$ $$ The RHS integral can be evaluated by substitution method. Let $u=mg+\beta v\;\rightarrow\;v=\dfrac{u-mg}{\beta}$ and $\;dv=\dfrac{du}{\beta}$ . Therefore $$ $\begin{aligned} h_f-h_i&=\frac{m}{\beta^2}\int_{v=v_f}^{v_i}\frac{u-mg}{u}\;du\\ &=\frac{m}{\beta^2}\int_{v=v_f}^{v_i}\left(1-\frac{mg}{u}\right)\;du\\ &=\frac{m}{\beta^2}\left.(u-mg\ln u)\right|_{v=v_f}^{v_i}\\ &=\frac{m}{\beta^2}\left.(mg+\beta v-mg\ln (mg+\beta v))\right|_{v=v_f}^{v_i}\\ &=\frac{m}{\beta^2}\left(\beta(v_i-v_f)-mg\ln\left(\frac{mg+\beta v_i}{mg+\beta v_f}\right)\right). \end{aligned}$ $$ Assuming the ball is thrown from the ground, we have $h_i=0$ and $\;v_f=0$ at the maximum height. Thus $$ $\begin{aligned} h_f-h_i&=\frac{m}{\beta^2}\left(\beta(v_i-v_f)-mg\ln\left(\frac{mg+\beta v_i}{mg+\beta v_f}\right)\right)\\ h_f-0&=\frac{0.1}{0.1^2}\left(0.1(10-0)-0.1\cdot9.8\ln\left(\frac{0.1\cdot9.8+0.1\cdot10}{0.1\cdot9.8+0.1\cdot0}\right)\right)\\ h_f&=10\left(1-0.98\ln\left(\frac{1.98}{0.98}\right)\right)\\ &\approx\boxed{3.10766\text{ m}} \end{aligned}$

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$\text{\# }\mathbb{Q.E.D.}\text{ \#}$

From energy conservation one can show that maximum height will be reached when velocity is zero.

Now analyzing the ball's motion along vertical, we have

$-ma\,=mg+\beta\,v$ where $a$ is ball's acceleration along vertical.

$\displaystyle$ $\rightarrow\,-v\times\frac{dv}{dh}\,=\frac{\beta\,v}{m}+g$

$\displaystyle$ $\rightarrow\,\frac{-m}{\beta}(1-\frac{mg}{\beta\,v+mg}) dv\,=dh$

Integrating within proper limits i.e. L.H.S. from $10$ to $0$ and R.H.S. from $0$ to $h_{max}$ and substituting the given values, we have

$h_{max}\,=10-(9.8)\times\,ln(\frac{10}{9.8}+1)\,=\boxed{3.108\;m}$ .

acceleration = $v+g$

$v dv/dx$ = $v+g$

$dx$ = $v/(v+g)$ $dv$

Integrate. Take v limit from 10 to 0 and x limit from 0 to x.

Starting from Newton's Second Law, we have $F=m \ddot z=-mg-\beta \dot z$ . Note the negative signs in front of both force terms; the first arises because gravity points downwards and the second because the drag acts in the opposite direction of the velocity.

Substituting the variable $v=\dot z$ for the velocity of the ball, and rewriting $\ddot z=\frac{dv}{dt}=\frac{dv}{dz}\frac{dz}{dt}=v\frac{dv}{dz}$ , we obtain the following:

$mv\frac{dv}{dz}=-mg-\beta v$

This can easily by solved by separation of variables. We then integrate from $0$ to $h_{max}$ for $z$ and from $v_i$ to $0$ for $v$ to calculate the maximum height of the ball.

$\int_{v_i}^0 \frac{v}{g+\frac{\beta}{m}v}dv=-\int_0^{h_{max}}dz$

Doing some manipulations on the improper fraction on the left hand side: $\frac{m}{\beta}\int_0^{v_i} (1-\frac{g}{g+\frac{\beta}{m}v})dv=\int_0^{h_{max}}dz$

Integrating: $h_{max}=\frac{mv_i}{\beta}-\frac{m^2g}{\beta^2}\ln\lvert1+\frac{\beta}{mg}v_i\rvert$

Finally, we substitute the values of $\beta$ , $m$ , $v_i$ and $g$ : $h_{max}=10-9.8\ln\lvert1+\frac{10}{9.8}\rvert=3.108m$

Newton’s second law $ma=~-\beta v-mg$

$\frac{dv}{dt}=-\frac{\beta }{m}v-g=-v-9.8$

$\frac{dv}{ds}\left( \frac{ds}{dt} \right)=-(v+9.8)$

$v\frac{dv}{ds}=-(9.8)$

-\underset{10}{\overset{0}{\mathop \int }}\,\frac{v}{v+9.8}dv=\underset{0}{\overset{s}{\mathop \int }}\,ds

$s\approx 3.11$

Just plug the value into this formula: $\frac{mv_{0}}{\beta}-\frac{m^{2}g}{\beta^{2}}\ln(\frac{g}{g-\beta v_{0}})$

where:

$m=0.1$

$g=-9.8$

$v_{0}=10$

$\beta=0.1$

You'll get $\boxed{3.10766439...}$ as the answer.