# Differential Equations

1. May 19, 2005

1) Show that since a body falling freely obeys the differential equation $$h''=-g[/itex], if it falls from an initial height $h(0)$, it lands with a velocity of $-\sqrt{2gh(0)}$ This problem is from a differential equations class and I solved it two different ways: Method 1: $\frac {dh} {dt}=v$ and $\frac {dv} {dt}=a$ which leads to $\int{a dh}=\int{v dv}$ and solving for v gives you $v=\sqrt{2gh}$ Method 2: Use $KE=PE$, so $\frac{1}{2}mv^2=mgh$ and solve for v which gives you $v=\sqrt{2gh}$. Unfortunately I don't think this is what the instructor is looking for. The wording of the question suggests he wants me to use differential equations techniques. I solved the DE to get $h=-\frac{1}{2}gt^2+c_1t+c_2$ but I'm not seeing how that will get me to $v=-\sqrt{2gh(0)}$ Can anyone help me? 2. May 19, 2005 ### dextercioby The velocity's equation v(t) is $v(t)=v_{0}-gt$ If u take the initial velocity to be zero and u found the falling time in terms of the height,then it's easy to get [tex]v\left(t_{\mbox{falling}\right)=-\sqrt{2g h(0)}$$

Daniel.

3. May 20, 2005

### HallsofIvy

Staff Emeritus
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.

4. May 21, 2005