Find the velocity of the falling object when it touches the water

Question

The question:

An object falls from a hovering surf-lifesaving helicopter over a port at $500$ m above sea level. Find the velocity of the object when it hits the water when the acceleration of the object is $0.2v^2 − g$. (Note that $g \not= 9.8$, because they did not state anywhere.)

Since they gave us $a$ in terms of $v$, and a value for $x$, I used:

• $v\frac{dv}{dx} = 0.2v^2-g$
• $\frac{dv}{dx} = 0.2v - gv^{-1}$
• $\int\frac{1}{0.2v - gv^{-1}} dv=\int dx$
• $\frac{5}{2}\ln(\frac{1}{5}v^2-g)+c=x$

How do I find the value of $c$, given only one info and where do I go next, after that?

Answer 1

Hint.

$$ -\dot v = \mu v^2-g, \ v(0) = 0, \Rightarrow v = \sqrt{\frac{g}{\mu}}\tanh\left(\sqrt{\mu g}t\right) $$

but

$$ \dot h = \sqrt{\frac{g}{\mu}}\tanh\left(\sqrt{\mu g}t\right),\ \ h(0) = h_0 $$

so

$$ h = h_0 + \frac{\ln\left(\cosh\left(\sqrt{\mu g}t\right)\right)}{\mu} $$

and $h=0$ for $t = \frac{\cosh^{-1}\left(e^{h_0\mu}\right)}{\sqrt{\mu g}}$

hence

$$ v = \sqrt{\frac{g}{\mu}}\left(1+e^{-h_0\mu}\right)\sqrt{\tanh\left(\frac{h_0\mu}{2}\right)} $$

and after substitution of $h_0=500, g = 9.8, \mu = 0.2$ in due units, we obtain

$$ v = 7[m/s] $$

Answer 2

First you find the path $x(t)$ along which our object moves. For this, simply solve the differential equation $\dot v=v^2-g$. You can solve this by separation of the variables using partial fraction decomposition of the integrand. Now you have a function $v(t)$, presumably with a free parameter. Since the object falls from a hovering object, its velocity at $t=0$ is $0$. So use the free parameter to enforce $v(0)=0$. You now have $v(t)$. Integrate it to obtain $x(t)$, again with a free parameter. Use this free parameter to enforce $x(0)=500\mathrm m$, since that's the height where it starts falling. Now find the point in time where $x(t)=0$ and find the corresponding velocity.

Answer 3

Considering that $500$ is a large distance for the object to reach an eventual velocity that renders the effective acceleration zero, i.e. $0.2v^2 - g =0$, which leads to

$$v= \sqrt{\frac g{0.2}}=\sqrt{5g}=7\text{m/s}$$