# Upward projectile motion differential equations

So I have this question here which says:

"An object of mass m is thrown vertically upward from the surface of the earth. It is subject to a constant gravitational field and air resistance proportional to the square of the object's velocity. If $g > 0$ is the constant acceleration due to gravity and $b > 0$ is the coefficient of quadratic air resistance, then Newton's second law gives

$\frac{dv}{dt}=-g-\frac{b}{m}v^2$;

where v(t) = $\frac{dz}{dt}$ is the object's velocity as a function of time $t= 0$. (Here we assume that when the object is moving upward, its velocity is positive). Solve the differential equation to find $v(t)$ and the height $z(t)$ of the object as it travels upward. Find the maximum height of the object, if its initial velocity is $v_0 > 0$."

This is obviously separable and I get

$\frac{dv}{-g-\frac{b}{m}v^2}=dt$

Which I feel I did correctly. I should then be able to integrate and then solve...

If I integrate, I get...

$\int_{v_o}^v \frac{dv}{-g-\frac{b}{m}v^2}$ = $\int_{0}^t dt$

I assume my bounds are correct. If I try to integrate tough, I get a really nasty substitution problem and end up with arc tan so I'm a little skeptical of if that's the right process...

Any guidance?

• If you pull a $-g$ from the denominator this substitution is actually pretty straightforward, it's actually just proportional to $\arctan{v}$ Commented Jan 23, 2017 at 18:29
• That is $\int \frac{dv}{-g-\frac{b}{m}v^2}= -\sqrt{\frac{m}{bg}}\int \frac{du}{1+u^2}$ where the substitution $u=\sqrt{\frac{b}{mg}}v$ was made after a $-g$ was factored out of the denominator, the negative sign can be cancelled by changing the order of integration as $v_f =0$ as the answer noted. The above integral is Arctangent Commented Jan 23, 2017 at 19:26

Your way is true and answer is right. For an object of mass $m>0$ which is thrown vertically upward from the surface of the earth, and air resist proportional to the square of the object's velocity, with $b>0$ is the coefficient of quadratic air resistance, you obtain $$\int_{v_o}^v \frac{dv}{-g-\frac{b}{m}v^2}=\int_{0}^t dt$$ by integration you have $$\color{red}{\frac{-1}{g}\sqrt{\frac{mg}{b}}\arctan(\sqrt{\frac{b}{mg}}v)\Big|_{v_0}^v=t}$$ then object goes up untill stop (for a moment), in this case we have $v=0$ so $$\frac{-1}{g}\sqrt{\frac{mg}{b}}\Big(-\arctan(\sqrt{\frac{b}{mg}}v_0\Big)=t$$ and $$\color{blue}{t_{top}=\sqrt{\frac{m}{bg}}\arctan\Big(\sqrt{\frac{b}{mg}}v_0\Big)}$$

• Thank you very much! I just had to properly do the substitution to solve the ODE. Commented Jan 28, 2017 at 7:10

At the maximum height of the projectile, you have a velocity of $v = 0$. One thing you can do is:

$\hskip 6.5cm\int_{v_0}^0 \frac{dv}{-g-\frac{b}{m}v^2}dv = \int_0^tdt$

Here, $t$ will tell you the time at which the projectile reaches maximum height.

Then, I believe on the way down your differential equation should be:

$\hskip 7.1cm \frac{dv}{dt} = g - \frac{b}{m}v^2$

Your method so far is fully correct. Your bounds are placed correctly.

We have: $$\int_{v_0}^v \frac{dv}{-g-\frac{b}{m}v^2} = \int_{0}^t dt$$

$$-\int_{v_0}^v \frac{dv}{g+\frac{b}{m}v^2} = \int_{0}^t dt$$

Integrating both sides:

$$-\left[\sqrt{\frac{m}{bg}}\arctan\left(v\sqrt{\frac{b}{mg}}\right)\right]_{v_0}^v = t$$

You can substitute terms such as $\sqrt{\frac{b}{mg}}=k$ to make the problem easier to visualize and solve.

$$\left[\frac{mk}{b}\arctan(kv)\right]_{v_0}^v = -t$$

$$\left[k\arctan(kv)\right]_{v_0}^v = -\frac{bt}{m}$$

$$k\arctan(kv)-k\arctan(kv_0)= -\frac{bt}{m} \tag{1}$$

Rearranging everything appropriately should give you an explicit solution for $v(t)$, which should help answering your other questions.

The height $z(t)$ can be found by integrating $v(t)$ with respect to $t$.

The maximum height of the object will be when $v=0$, so set your expression equal to this. Conveniently, you could use the implicit form on equation $(1)$.

Hope this helps. If you have any doubts or questions, feel free to ask.