# What minimum speed must he be running?

1. Mar 16, 2008

### dflutedevil

A 17m--long vine hangs vertically from a tree on one side of a 10m--wide gorge, as shown in the figure. Tarzan runs up, hoping to grab the vine, swing over the gorge, and drop vertically off the vine to land on the other side. What minimum speed must he be running?

2. Mar 16, 2008

### hage567

What have you tried so far? You must show some work.

3. Mar 16, 2008

### dflutedevil

I just need some direction that's all..I'm not really good at this...I don't need the whole answer...just maybe some direction

4. Mar 16, 2008

### ReaverKS

Hi I'm new here, currently enrolled in Physics 101 (algebra based, but I'm in calc 3 right now) and I'm also curious as to how to start this problem. Thank you.

5. Mar 16, 2008

### Snazzy

This is a conservation of energy problem. Figure out how high Tarzan is from the base of the gorge at the end of his swing to give you your potential energy, from which you can calculate the initial kinetic energy, and thus his velocity.

6. Mar 16, 2008

### dflutedevil

well i got how high he was when he landed but how does that relate to potential energy. isn't that U=mgy. i dont have his mass? is there something I'm missing?

7. Mar 16, 2008

### Snazzy

You don't need the mass if you assume that all of the kinetic energy he had in the beginning is transferred into potential energy at the end of the swing.

8. Mar 16, 2008

### dflutedevil

I'm confused....I'm sorry. I guess I don't understand the concepts...so the h that i got = the potential and kinetic energy? but still the kinetic energy equation equals 1/2mv^2 which still requires mass?

9. Mar 16, 2008

### Snazzy

But if $$KE_i=PE_f$$ and you expand out what kinetic energy and potential energy equal, do you need the mass?

10. Mar 16, 2008

### dflutedevil

so is this right then...
1/2v^2=h thus....
v=sqrt(2h)

11. Mar 16, 2008

### Snazzy

Where did the g go?

12. Mar 16, 2008

### dflutedevil

uhm... don't know....so
v=sqrt(2gh)

13. Mar 16, 2008

### Snazzy

Looks better, now the hard part is finding the height above the ground at the end of the swing.

14. Mar 16, 2008

### dflutedevil

see i just used the triangle to find the hypotenuse and then subtracted 17 from that. is that how i was supposed to do it? or am i way off?

15. Mar 16, 2008

### ReaverKS

Is there some way you could either scan and upload the figure or possible explain the figure in words? I don't have enough information and I'm genuinely interested in the problem and solution.