Maximum height reached by a projectile

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
2 replies · 3K views
sid9221
Messages
110
Reaction score
0
http://dl.dropbox.com/u/33103477/gravity.png

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:
Physics news on Phys.org
sid9221 said:
http://dl.dropbox.com/u/33103477/gravity.png

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.

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

RGV
 
Last edited by a moderator: