# Find velocity given y and angle

1. Jun 22, 2010

### mandy9008

1. The problem statement, all variables and given/known data
Tarzan swings on a 34.0 m long vine initially inclined at an angle of 38.0° with the vertical.
(a) What is his speed at the bottom of the swing if he starts from rest?
(b) What is his speed at the bottom of the swing if he pushes off with a speed of 2.00 m/s?

3. The attempt at a solution
I honestly do not even know where to begin.

2. Jun 22, 2010

### wwshr87

Use conservation of energy, you know the height at which he starts, you know the height at the bottom of the swing. You know the initial speed for both cases. This should get you started. KEi + PEi = KEf + PEf

3. Jun 22, 2010

### inky

(1)Firstly draw freebody diagram. Consider how many forces acting on Tarzan.
(2) Use Newton's Second law then find acceleration.
(3) Find the length between tarzan initial and final condition.
(4) Find the velocity at the bottom.

4. Jun 22, 2010

### mandy9008

KE + PE = KE + PE
mgh = 1/2 mv2
v2 = gh/2
v2 = (9.8 m/s2)(34m) / 2
v= 12.91 m/s

where does the angle come into play?
when i entered this answer, it told me that I was within 10% of the correct answer

5. Jun 22, 2010

### wwshr87

See figure attached.

#### Attached Files:

• ###### angle.JPG
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6. Jun 22, 2010

### joshmdmd

basically find the change in height using trig... the length of the vine * cos(angle)

then you take the length and subtract what you found..
next find (mass)(9.81)(change in height)

equate that to 1/2mv^2

so take the answer, multiply by 2, divide by mass and then get the squareroot..

thats the velocity..

all this is is mgh=1/2mv^2 (Eg=Ek)

7. Jun 22, 2010

### mandy9008

I don't know the mass though

8. Jun 22, 2010

### wwshr87

mandy, if you write out KEi + PEi = KEf + PEf, you will notice a common term which can be canceled.

9. Jun 22, 2010

### joshmdmd

i just did the question.. ill post the solution.. scanning it atm..

mass cancels out.. are the answers 16 and 16.1m/s?

10. Jun 22, 2010

### mandy9008

I did cancel out the mass when I solved this equation. Is this not the correct equation?

11. Jun 22, 2010

### joshmdmd

its not bc you're dividing by 2 when you should be multiplying.. you're also ignoring the angle and delta heighthttp://yfrog.com/6bscan0001yj

12. Jun 22, 2010

### joshmdmd

http://img227.imageshack.us/img227/5659/scan0001y.jpg [Broken]

this is my solution..

Last edited by a moderator: May 4, 2017
13. Jun 22, 2010

### wwshr87

The equation is correct, and you canceled out the mass, but the height is not correct. Instead use delta h in the figure which i attached. Tarzan did not release from 0 degrees, therefore take this into account.

14. Jun 22, 2010

### wwshr87

joshmdmd, you must be rather careful here. If the 38 degrees are measured from the x-axis like I have drawn then in order to obtain the y-component you must take the sine of the angle.

15. Jun 22, 2010

### joshmdmd

16. Jun 22, 2010

### joshmdmd

mmm

careful?

cos(90-38) = sin38

17. Jun 22, 2010

### joshmdmd

oops.. i took theta in your drawing as 38.. =] you messed me up

18. Jun 22, 2010

### wwshr87

Sorry for the confusion, just making sure you know whats going on.

19. Jun 22, 2010

### joshmdmd

here we go.. corrections to answer..

v=(SQRT)2gh
v=11.886m/s

b)
v=12.22

20. Jun 22, 2010

### mandy9008

Josh, the answer in your scan is wrong. When i did it and got 12.91 m, it said that I was within 10% of the correct answer.