# Maximum height reached by a projectile

1. May 15, 2012

### sid9221

http://dl.dropbox.com/u/33103477/gravity.png [Broken]

I have worked out the first two bits. (So you can assume that I have that) But I can't figure out how to work out the maximum height reached.

I know there I can equal the KE and the PE to work out max height, but that doesn't look like it'll work here.

Last edited by a moderator: May 6, 2017
2. May 15, 2012

### vela

Staff Emeritus
Integrate v(t) to find y(t).

3. May 15, 2012

### Ray Vickson

The KE/PE argument will not work because of the presence of air drag. However, you don't need it. You have $v = dy/dt.$ Do you recall the conditions for a maximum of y(t)?

RGV

Last edited by a moderator: May 6, 2017