Solving Motion Problems: John and Pedro

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Homework Help Overview

The discussion revolves around two motion problems involving a window washer named John and a character named Pedro. The first problem involves calculating the time it takes for a dropped squeegee to reach the ground from a height of 29.4 meters. The second problem concerns determining how far a ball, tossed horizontally from a height of 78.4 meters, lands from the base of the building.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the appropriate formulas for vertical and horizontal motion, with some attempting to apply kinematic equations. Questions arise regarding the initial conditions, such as whether the vertical velocity is zero when the ball is tossed and how to handle horizontal and vertical components separately.

Discussion Status

Some participants have made progress on the first problem, with one confirming a time of 3 seconds for the squeegee drop. The second problem remains less resolved, with ongoing questions about vertical displacement and the use of horizontal velocity. Guidance has been offered regarding the separation of horizontal and vertical motions, but no consensus has been reached on the final answers.

Contextual Notes

There is some confusion regarding the initial vertical velocity and the application of acceleration in the horizontal direction. Participants are also navigating the constraints of homework rules and the need for clarity in the problem setup.

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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
 
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Give it a shot. What formulas do you know? When the squeegee is dropped, does it accelerate?
 
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.
 
tdusffx said:
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. :smile:
 
hmmm..no luck on 2nd one...can you help me on this? lol
 
tdusffx said:
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.
 
I have the vertical displacement, right? isn't the vertical displacement 78.4m?
 
tdusffx said:
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.
 
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.
 
  • #10
tdusffx said:
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?
 
  • #11
I got 6.5secs for the vertical..i'm working on the horizontal.
 
  • #12
how do I find the horizontal velocity?
 
  • #13
tdusffx said:
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?
 
  • #14
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?
 
  • #15
tdusffx said:
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 ?
 
  • #16
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
tdusffx said:
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
 
  • #18
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
tdusffx said:
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?
 
  • #20
oh ya...Xi is 0...sorry about that

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

Xf = 158.5?
 
  • #21
tdusffx said:
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?
 
  • #22
i don't know what to use for a...

:(
 
  • #23
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.
 
  • #24
tdusffx said:
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.
 
  • #25
tdusffx said:
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.
 

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