# Homework Help: Motion problems

1. Sep 17, 2007

### tdusffx

1.) john, a window washer, is 29.4 m high on a scaffold rising at a speed of 4.9 m/s when he drops a squeegee. ignoring air resistance, the time it takes the squeegee to reach the ground is?

2.) pedro tosses a ball out of a window 78.4m above level after checking to see that no one on the ground might be hit by the ball. if the ball exits the window with the horizontal velocity of 20m/s, how far from the point below the window does it land?

I dont know which formula is to use in these problems..help plz =D

2. Sep 17, 2007

### learningphysics

Give it a shot. What formulas do you know? When the squeegee is dropped, does it accelerate?

3. Sep 17, 2007

### tdusffx

i solved the first one...I used the xf = Xi +Vxi + 1/2at^2 formula for the first one..

2nd one i'm still trying.

4. Sep 17, 2007

### learningphysics

Cool. That's the right formula to use, but the formula should be:

xf = Xi +Vxi*t + 1/2at^2

probably just a typo.

good.

5. Sep 17, 2007

### tdusffx

hmmm..no luck on 2nd one...can you help me on this? lol

6. Sep 17, 2007

### learningphysics

sure. can you give an equation for vertical displacement? I assume that the scaffold isn't still rising at this point... the question doesn't make it clear.

7. Sep 17, 2007

### tdusffx

I have the vertical displacement, right? isn't the vertical displacement 78.4m?

8. Sep 17, 2007

### learningphysics

That's the initial height... try to use this equation just as before:

xf = Xi +Vxi*t + 1/2at^2

only difference is that Vxi = 0 I believe.

9. Sep 17, 2007

### tdusffx

hmm..I think Xf = Xi + Vit +1/2at^2

you just change it to x's to y's since your doing vertical...

therefore..0 = 78.4 + 20x + 1/2(-9.81)t^2

I got up there...I dont kow what to do next.

10. Sep 17, 2007

### learningphysics

You should deal with the horizontal motions and vertical motions separately. For the vertical equation, you use only vertical velocity and vertical acceleration.... for the horizontal equation, you use only horizontal velocity and horizontal acceleration.

What answer did you get for the first part?

11. Sep 17, 2007

### tdusffx

I got 6.5secs for the vertical..i'm working on the horizontal.

12. Sep 18, 2007

### tdusffx

how do I find the horizontal velocity?

13. Sep 18, 2007

### learningphysics

Is that what you got for part 1) of the question? That's not what I'm getting. Can you post your calculations?

14. Sep 18, 2007

### tdusffx

78.4 + 20t + 1/2(-9.81)t^2 = 0

t = 6.5, -2.4

then for the x components

Yf = 0 + Vxi(6.5) + 1/2(-9.81)(6.5)^2

But i dont know how to find Vxi...is it 20m/s also?

15. Sep 18, 2007

### learningphysics

You shouldn't be using 20 here... 20m/s is the horizontal velocity not vertical. What is the initial vertical velocity?

What did you get for the previous part of the question... the one with 29.4m going up wat 4.9m/s ?

16. Sep 18, 2007

### tdusffx

i got 3 seconds for the first part of the question...so like, use the 20m/s for the horizontal..and the vertical? :D

17. Sep 18, 2007

### learningphysics

Ah good. I got 3s also. The initial vertical velocity is 0 (assuming the scaffold is not moving up at this point). So for the vertical part your equation should just be:

78.4 + 1/2(-9.81)t^2 = 0

18. Sep 18, 2007

### tdusffx

oh kk...

78.4 + 1/2(-9.81)t^2 = 0

t = plus/minus 4

therefore

Xf = 78.4 + 20(4) 1/2(-9.81)(4)^2?

Xf = 79.9 meters?

19. Sep 18, 2007

### learningphysics

Remember for the horizontal equation... you use horizontal velocity... horizontal acceleration etc... why are you using -9.81 for the horizontal displacement? same with the 78.4... ?

what is the horizontal acceleration? what is the initial horizontal position?

20. Sep 18, 2007

### tdusffx

oh ya...Xi is 0...sorry about that

Xf = 20(4) + 1/2(9.81)(4)^2

Xf = 158.5?