Show velocity of a particle during its flight at time $t$

Question

I'm completely stuck, I think I have to use Newton's second law but I have no idea where to start, any help would be appreciated!

At time $t=0$ a particle of unit mass is projected vertically upward with velocity $v_0$ against gravity, and the resistance of the air to the particle's motion is $k$ times its velocity. Show that during its flight the velocity $v$ of the particle at time $t$ is:

$$v = \left(v_{0} + \frac{g}{k}\right) e^{-kt} - \frac{g}{k}$$

Deduce that the particle reaches its greatest height when

$$t = \frac{1}{k} \ln\left({1+\frac{kv_{0}}{g}}\right)$$

and that the height reached is

$$ \frac{v_{0}}{k} - \frac{g}{k^2} \ln{\left(1 + \frac{kv_{0}}{g}\right)}$$

Thanks!

Answer 1

After making edits, my original comment was too long. So, I've made it into an answer. Note: I had deleted my answer, but I just undeleted for more of a hint to the problem rather than a full solution (as well as for the few sentences following this.)

What were your forces acting on the ball? When you're looking to write a differential equation for an introductory physics course (involving air resistance or for SHM), you should immediately think of an equation involving net force or one that involves net torque ($\sum F = {ma}$ or $\sum \text{torque} = I\alpha$). Obviously in this case you want to deal with forces.

To expand, once you write your net force statement $\sum F = ma$, you can then rewrite $a$ as $\frac{dv}{dt}$ and simply separate variables and integrate both sides. Note that the mass is considered $1$ so your integration is made slightly easier. It will take some "prettying up" after you integrate. Also, a hint for the second part, when is a key feature when something is at its max height even for just an instant?

Answer 2

Newton's second law isn't really important here; it's only used in calculating the acceleration caused by drag, and the particle is unit mass so it's exactly the force caused by drag. Once you've worked that out, you can forget about Newton's second law altogether; the real substance of the question is purely in the following facts:

• acceleration is the derivative of velocity, and so velocity is the integral of acceleration
• velocity is the derivative of distance, so distance is the integral of velocity.

The effect of gravity is given in terms of acceleration, and the air resistance is easily converted to acceleration, so you can formulate an equation between acceleration and velocity, and then turn it into a differential equation using the first bullet point, which can be solved by separation of variables.

At the point of maximum height, the particle must neither be going up nor coming down. Translate that into a condition on the velocity and then plug it in your equation.

To calculate the height reached, use the second bullet point.

Answer 3

Start off by writing the equation:

$$\sum F = ma$$

$$ma=F_d +mg$$

where $F_d=kv$ is the drag force. and the particle is given to be of unit mass so the equation becomes.

$$-\frac{dv}{dt}=kv+g$$

$a=-\frac{dv}{dt}$ since velocity is decreasing,taking similar terms on same side and integrating with proper limits, we get

$$\int_{v_0}^v\frac{dv}{kv+g}=\int_o^t -dt$$

$$\frac{\ln(kv+g) - \ln(kv_0+g)}{k}=-t$$

$$\ln{\frac{kv+g}{kv_0+g}}=-kt$$

Solving for $v(t)$ gives us $$v = (v_{0} + \frac{g}{k}) e^{-kt} - \frac{g}{k}$$

To find the time when max height is reached, put $v=0$, and find the corresponding value of $t$(say $t=t_0$)

$$(v_{0} + \frac{g}{k})e^{-kt_0}=\frac{g}{k}$$ this gives

$$t_0=\frac{1}{k} \ln\big({1+\frac{kv_{0}}{g}}\big)$$

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To find max height

write $v = (v_{0} + \frac{g}{k}) e^{-kt} - \frac{g}{k}$ as,

$$\frac{dx}{dt} = (v_{0} + \frac{g}{k}) e^{-kt} - \frac{g}{k}$$ Integrating with proper limits, $$\int_0^h dx=\int_o^t (v_{0} + \frac{g}{k}) e^{-kt}.dt - \int_0^t\frac{g}{k}.dt$$

$$h=(v_0+\frac{g}{k})\frac{e^{-kt}}{-k} -(v_0+\frac{g}{k})\frac{1}{-k} -\frac{gt}{k}$$

putting $t=t_O$ in this equation to get max height,(since max height occurs when $v=0$, which happens when $t=t_0$)

$$ h_{max}=\frac{v_{0}}{k} - \frac{g}{k^2} \ln{(1 + \frac{kv_{0}}{g})}$$