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 h&#039;&#039;=-g[/itex], if it falls from an initial height h(0), it lands with a velocity of -\sqrt{2gh(0)}<br /> <br /> This problem is from a differential equations class and I solved it two different ways:<br /> <br /> <u><b>Method 1</b></u>:<br /> \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}<br /> <br /> <u><b>Method 2</b></u>:<br /> Use KE=PE, so \frac{1}{2}mv^2=mgh<br /> and solve for v which gives you v=\sqrt{2gh}.<br /> <br /> Unfortunately I don&#039;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&#039;m not seeing how that will get me to v=-\sqrt{2gh(0)}<br /> <br /> 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&#039;(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