Using Differential Equations to Solve for Velocity of a Free Falling Body

[mekkomhada]

1) Show that since a body falling freely obeys the differential equation [tex]h''=-g[/itex], if it falls from an initial height $h(0)$, it lands with a velocity of $-\sqrt{2gh(0)}$

This problem is from a differential equations class and I solved it two different ways:

Method 1:
$\frac {dh} {dt}=v$ and $\frac {dv} {dt}=a$ which leads to $\int{a dh}=\int{v dv}$ and solving for v gives you $v=\sqrt{2gh}$

Method 2:
Use $KE=PE$, so $\frac{1}{2}mv^2=mgh$
and solve for v which gives you $v=\sqrt{2gh}$.

Unfortunately I don't think this is what the instructor is looking for. The wording of the question suggests he wants me to use differential equations techniques. I solved the DE to get $h=-\frac{1}{2}gt^2+c_1t+c_2$ but I'm not seeing how that will get me to $v=-\sqrt{2gh(0)}$

Can anyone help me?

 

[dextercioby]

The velocity's equation v(t) is $v(t)=v_{0}-gt$ If u take the initial velocity to be zero and u found the falling time in terms of the height,then it's easy to get

$$v\left(t_{\mbox{falling}\right)=-\sqrt{2g h(0)}$$

Daniel.

 

[HallsofIvy]

Actually, your "method 1" is a differential equation method.

However, you can, as you say get $$h(t)= -\frac{1}{2}gt^2+ c_1t+ c_2$$ and then $$v(t)= h'(t)= -gt+ c_1$$.
Taking t= 0 to be the moment the body is dropped, h(0)= c_2 and v(0)= c_1= 0 so
$$h(t)= -\frac{1}{2}gt^2+ h(0)$$ and $$v(t)= -gt$$.

The body "lands" when h(t)= 0. Solve $$h(t)= -\frac{1}{2}gt^2+ h(0)= 0$$, which is thesame as $$t^2= \frac{2h(0)}{g}$$, (of course, you only want the positive root) and put into v(t)= -gt to find speed with which it lands.

 

[mekkomhada]

Thanks for helping me out...I don't know why, but the answer wasn't immediately apparent to me. I need a break :tongue: