A roller coaster, shown in Fig. 8-29 (h1 = 40 m, h2 = 16 m, h3 = 28 m), is pulled up to point A where it and its screaming occupants are released from rest. Assuming no friction, calculate the speed at points B, C, D

I got the equation mgh+.5v2=mgh=.5v^2 and i keep getting it wrong.



A 75 kg trampoline artist jumps vertically upward from the top of a platform with a speed of 3.5 m/s.


How fast is he going as he lands on the trampoline, 2 m below?
m/s

( If the trampoline behaves like a spring of spring constant 5.2 x 10^4 N/m, how far does he depress it?
m

Hi, as with your other problem it should be mgh=1/2*mv^2 but the h in question is the difference in heights between points A and B, B and C, and C and D.
I can't see the picture but suspect that A is 40m above B which is 16m below C which itself is 28m above D.
In this case by point B, 1/2*mv^2 = 40mg or v^2= 80g so v will be about 28m/s. If you apply the same process to the next stage of the ride you should notice a deceleration of 40% (due to going up 40% of initial drop) - and this must be ADDED to the inital speed (28ish) to get the speed at point C. Alternatively you can think of the first two sections of track as one larger section and do your mgh=1/2*mv^2 for the overall. This is probably the best way to check you figures so do it a section at a time and then do the whole thing.
If I have got the shape all wrong or if this is no help then try the Mechanics section of:

http://forum.physorg.com/index.php?showtopic=3761

The second part is again about the artist turning his initial kinetic energy into potential, and it then being turned back into kinetic during his downward motion.
Because the mass is a factor of both equations it can be ignored to begin with hence
gh = 1/2*v^2 so h = 12.25/2g or .6m ish up from initial point. Now you need to work out speed after (2m + answer a) drop.