Posted by: xXIHAYD0IXx
Posted by: TheDerRitter
Posted by: xXIHAYD0IXx
Don't worry OP, this problem isn't that bad. The second equation you have, though, is the potential energy of a spring, so I'd through that out. You shoul know how heavy the board is, so I think maybe you wrote the problem wrong, but that's okay, as mass cancels out anyway. All you do is set:mgh=.5mv^2So if the distance h is 5 meters (when the board has fallen half way), your velocity should be sqrt(2gh). When it falls all the way, h becomes 10, not five. So, ultimately, your two answers should be 9.9m/s (like you said) and 14m/s. It's as easy as that. I knew college taught me something. Let me know if you have any questions.How is it 10? It says it's falling so it should be 0 when it hits.I edited my last post with the answers. Just an FYI. The "h" in the equation is the initial height above the surface from which the object is dropped, not the final height. Technically, the equation should read mgh_i + .5m(v_i)^2 = mgh_f + .5m(v_f)^2From this, it should be easy to see that v_i=0, as the object is initially at rest and h_f is zero, as the object is just barely above the water. Thus, those two terms go to zero, leaving you with mgh_i = .5m(v_f)^2EDIT- Thank you for correcting the problem, but as I said, mass cancels out. Also, you don't have to trust me if you don't want, but I just took a college physics final over (roughly) the same material two days ago, so it is quite fresh in my mind and I assure you this is right. One more thing, annoyinginge is technically right, though his method is a bit more convoluted.
Alright, I got it, just a few more tests and I'll report back.
Alright, so I think I get it now: H isn't 0, it's 10 since it's the initial and the final is 0. The final would be thrown out, leaving the initial.
With that: mgh_i= (77kg)(9.8m/s)(10m) which would make it 7550J, and since 1/2(77kg)v^2, M for 1/2mv^2 would be 38.5kg
and then:
7550J/38.5=196.1038961 and the sqrt of that would be 14m/s!
[Edited on 12.10.2012 10:15 AM PST]