Intro to Differential Equations

As the ball goes up, the equation of motion reads:

$\frac{dv}{dt} = -g-av$

Where $a = \frac{k}{m}$

Separating the variables and integrating, knowing that at $t=0$ , $v=v_1$ , gives:

$v = \frac{(g+av_1)e^{-at}-g}{a}$

When the ball reaches the maximum height, let us assume that a time $t_1$ has elapsed, and $v=0$ . Using this fact in the above solution yields:

$(g+av_1)e^{-at_1}=g$

Now $v = \frac{dx}{dt}$

Solving for $x(t)$ involves integrating the velocity:

$\int_{0}^{x}dx = \int_{0}^{t} v dt$

Gives:

$x = \frac{(g+av_1)(1-e^{-at})}{a^2}-\frac{gt}{a}$

Let us say that the maximum height is $h$ . The maximum height is reached at time $t_1$ . This gives:

$h = \frac{(g+av_1)(1-e^{-at_1})}{a^2}-\frac{gt_1}{a}$

Using the fact that $(g+av_1)e^{-at_1}=g$ in the equation above gives:

$h = \frac{v_1-gt_1}{a}$

Now, the motion is analysed as the ball moves downward. The equation of motion reads:

$\frac{dv}{dt} = g-av$

When $t=0$ , then $v=0$ . Separating variables and integrating gives the solution:

$v = \frac{g}{a}\left(1-e^{-at}\right)$

The downward distance $x$ moved can be obtained by integrating the velocity again, giving

$x = \frac{g}{a}\left(t - \frac{(1-e^{-at})}{a}\right)$

The ball hits the ground at time $t_2$ and has moved a distance $h$ and acquires a downward speed of $v_2$ . Using all this gives us the relation for $h$ in the same manner as obtained for when the ball moves up. The equation is:

$h = \frac{-v_2+gt_2}{a}$

We already obtained:

$h = \frac{v_1-gt_1}{a}$

Subtracting these two equations and simplifying gives us the required answer:

$t_1 + t_2 = \frac{v_1 +v_2}{g}$

The solution has quite a few step jumps. The problem is manageable but it is a bit tedious.