# X and y component problem?

1. Feb 12, 2008

### Lacy

[SOLVED] x and y component problem?

1. The problem statement, all variables and given/known data

http://loncapa3.physics.sc.edu/cgi-bin/plot.png?file=dengjh_sc_1202847653_17793457_plot.data [Broken]

u will need that graph to solve this problem

) A golf ball is struck at ground level. The speed of the golf ball as a function of the time is shown in the figure., where t = 0 at the instant the ball is struck. How far does the golf ball travel horizontally before returning to ground level? i found this to be 112m and it is right

What is the maximum height above ground level attained by the ball?

2. Relevant equations
S = (1/2)*a*t^2

3. The attempt at a solution

Looks like the ball travels for 4 sec. When the speed is minimum (that speed looks like 27.5 m/s to me) it has reached its peak so the vertical component of speed has gone to zero. All the remaining velocity is horizontal, 27.5 m/s. And that's the horizontal velocity,Vh, during the entire 4 seconds. So the horizontal distance h is found from
Vh = h / total time

Vertical distance: It took 2 seconds to return to the ground from that peak,
S = (1/2)*a*t^2

now i tried using the equation but it didnt work. can someone point me to the right directioin

Last edited by a moderator: May 3, 2017
2. Feb 12, 2008

### Hootenanny

Staff Emeritus
Welcome to PF Lacy,

3. Feb 12, 2008

### Lacy

thank you i fixed it

4. Feb 12, 2008

### Hootenanny

Staff Emeritus
The reason it didn't work is your equation is incomplete. The full equation is,

$$x = x_0 + v_0 t + (1/2) a t^2$$

Where x0 is the initial displacement (in this case zero) and v0 is the initial velocity.

5. Feb 12, 2008

### Lacy

now im trying to find max height would i use y=vot-.5at^2

would this be correct

y=34(2)-(4.9)(4)

48.4

Last edited: Feb 12, 2008
6. Feb 12, 2008

### Hootenanny

Staff Emeritus
Correct, where a = 9.81m.s-2

7. Feb 12, 2008

### Lacy

would this be correct

y=34(2)-(4.9)(4)

48.4

8. Feb 12, 2008

### Hootenanny

Staff Emeritus
How did you determine the initial velocity?

9. Feb 12, 2008

### Lacy

i looked on the graph and it started at 34m/s

10. Feb 12, 2008

### Lacy

i also got this equation but its different from the first one

you rearrange y-yo = Vosin(theta)t-
1/2gt^2 so that sin(theta) = (1/2gt^2)/Vo. Once you have
sin, plug in the variable in the y-yo = Vosin(theta)t-
1/2gt^2 formula.

which way do i do it im confused

11. Feb 12, 2008

### Hootenanny

Staff Emeritus
That is the initial speed, not the initial [vertical] velocity. However, you know the horizontal velocity (112/4) and the speed, therefore you can easily calculate the initial vertical velocity.

12. Feb 12, 2008

### Lacy

y=28(2)-(4.9)(4)

alright now i got 36.4 is this right
where vo=28 and t=2

13. Feb 12, 2008

### Hootenanny

Staff Emeritus
[COLOR="#black"]White space[/COLOR]

14. Feb 12, 2008

### Lacy

would this be it

vo=.5at^2

where a=9.8 t=2

so vo=19.6m/s ???

15. Feb 12, 2008

### Hootenanny

Staff Emeritus
You've use the wrong equation, but got the correct answer. The correct method would be as follows,

$$v = v_0 + at$$

At maximum height $v=0$, hence,

$$v_0 + at = 0 \Rightarrow v_0 = -2g = 19.6 m.s^{-2}$$

16. Feb 12, 2008

### Lacy

alright i got the right answer THANK YOU for all the help hootenanny :)

17. Feb 12, 2008

### Hootenanny

Staff Emeritus
A pleasure Lacy

If you'd be so kind to mark the thread as 'solved' (look for "Thread Tools" in the top right-hand corner of the thread) that would be great.