Solving Motion Problems: John and Pedro

[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 don't know which formula is to use in these problems..help please =D

 

[learningphysics]

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

 

[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.

 

[learningphysics]

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

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

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

probably just a typo.

2nd one I'm still trying.

good.

 

[tdusffx]

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

 

[learningphysics]

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

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.

 

[tdusffx]

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

 

[learningphysics]

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

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.

 

[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 don't kow what to do next.

 

[learningphysics]

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 don't kow what to do next.

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?

 

[tdusffx]

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

 

[tdusffx]

how do I find the horizontal velocity?

 

[learningphysics]

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

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

 

[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 don't know how to find Vxi...is it 20m/s also?

 

[learningphysics]

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

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 ?

 

[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

 

[learningphysics]

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

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

 

[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?

 

[learningphysics]

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?

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?

 

[tdusffx]

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

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

Xf = 158.5?

 

[learningphysics]

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

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

Xf = 158.5?

why are you using 9.81?

 

[tdusffx]

i don't know what to use for a...

:(

 

[tdusffx]

well, i have to sleep now..thanks for the help sir. I have physics 7:30 (central time) am tomorrow, lol. have a great night and thanks again.

 

[learningphysics]

i don't know what to use for a...

:(

a = 0. There is no acceleration horizontally. Acceleration only happens when there's a force... the only force here is gravity which causes a vertical acceleration only.

in other words... the velocity is constant horizontally.

 

[learningphysics]

well, i have to sleep now..thanks for the help sir. I have physics 7:30 (central time) am tomorrow, lol. have a great night and thanks again.

no prob. good night.