Recent content by Finn007

Ohhh, okay. So I'm doing first: vf2 = sqrt(2gh) = sqrt(2(9.8)(3)) = 7.7 for finding the velocity when they land at the other side. Then use V = 7.7: v1i = ((m1+m2)V)/m1 = ((66+25)/7.7)/66 = 10.6 for finding the velocity just before the rescuer reaches the dog. Then use v = 10.6 h = v2/2g =...

Can I assume that mv2/2 = mgh and v is v1i?

So how would I connect finding v1i to the height? Should I use KE = mv2/2 and GPE = mgh?

Inelastic? But I'm not sure which equations to use that could help me find the height. I was thinking m1v1i = (m1+m2)V

I think that energy is conserved based on my assumption but I'm not sure.

hl = (((mr+md)h2)/mr)-h1 = (((66+25)3)/66)-5 = -0.8636

25.0-kg dog is trapped on a rock in the middle of a narrow river. A 66.0-kg rescuer has assembled a swing with negligible mass that she will use to swing down and catch the trapped dog at the bottom of her swing, and then continue swinging to the other side of the river. The ledge that the...