Hello all,

I have a little problem solving these 2 problems, guess you all can help me cuz its piece

of cake for you all.

Ok here we go:

1. Old faithful geyser in yellowstone national park shoots water every hour to a height of

40.0m. With what velocity does the water leave the ground?

2. A pendulum with a mass of 1 kg is released from a height of 1.5cm above the height of

its restin position. How fast will the pendulum be moving when it passes thorugh the

lowest point of its swing?

Dunno why when I saw this kind of problem(like number 1) ,only giving 1 value, is kinda

to solve for me..

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Doc Al
Mentor
To get detailed help, please show your work and point out where you got stuck.

Hint: Both problems can be solved using conservation of energy.

Errn.. Well, I cannot post any link cuz I dont have 15 posts nor image cuz it exceeds the

required size.

I can tell the answers, 1. Vf= 28m/s

2. Vf=0.55m/s

Thats what I have at the moment, is it correct?

Doc Al
Mentor
Show how you got those answers. (We don't need to see a diagram.)

PE = KE

vf^2= 2mgh/m=2gh

Vf^2=2*40*9.81

Vf^2=784.8

Vf=28.0m/s

----------------

PE = KE

I changed 15cm into m so it will be = 0.15

PE=mgh

PE=1(9,81)(0.15)

PE=1.5

KE=1/2mgv^2

KE= 1/2(1)

KE= 0.5

KE/PE= Vf^2

0.5/1.5=Vf^2

0.3=Vf^2(Apply square root to both side)

0.55=Vf

I got it this way.. tell me if Iam wrong

Nabeshin