Romeo's Pebbles - A Zero Horizontal Acceleration Problem

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Homework Help Overview

The problem involves projectile motion, specifically analyzing the motion of pebbles thrown from a distance to hit a target at a certain height. The original poster is tasked with determining the speed of the pebbles upon impact with a window, given specific vertical and horizontal distances.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • The original poster lists knowns and unknowns and attempts to apply kinematic equations to find the initial velocity and hang time. Some participants question the validity of the equations used and suggest a two-dimensional approach to the problem.

Discussion Status

Participants have provided feedback on the original poster's approach, indicating errors in the equations used and suggesting a need for a more comprehensive analysis of the projectile motion. There is acknowledgment of a corrected formula leading to a different result, but no consensus on the overall method has been reached.

Contextual Notes

The original poster expresses urgency due to a looming deadline and indicates frustration with the problem-solving process. There is a mention of a specific answer being achieved after corrections, but the details of the corrections are not fully explored.

EcKoh
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Homework Statement



Romeo is chucking pebbles as gently as he can up to Juliet's window. That is, he wants the pebbles to hit the wendow with the least possible speed. He is standing at the edge of a rose garden at Dv = 7.90 m below her window and at Dh = 8.50 m from the base of the wall. How fast are the pebbles going when they hit her window?

Homework Equations



Vf = Vi + adv (to find initial velocity)
Vf = Vi + at (to find hang time)
Dh = Vht (to find constant horizontal velocity)


The Attempt at a Solution



Okay so I'm assuming I did everything right, so I'm not sure where I went wrong but here are the steps I took in my attempt:

3.1: List knowns and unknowns:

Knowns:
Vf = 0 m/s
a = -9.8 m/s2
Dv = 7.90 m
DH = 8.50 m

Unknowns:
Vi = ?

3.2: Find the initial velocity:

Vf = Vi + adv
(0) = Vi + (-9.8)(7.90)
Vi = 77.42

3.3: Find hang time:

Vf = Vi + at
(0) = (77.42) + (-9.8)t
-77.42 = -9.8t
t = 7.9s <-- (maybe I went wrong here? I noticed that's the same as Dv

3.4: Find constant velocity:

Dh = Vht
(8.50) = Vh(7.9)
Vh = 1.08

So 1.08 m/s turns out to be the wrong answer, and I keep going back and reworking it but I'm not sure what I'm doing wrong. Its due online tonight in just a few hours, but either way it's driving me crazy so I just want to know what I'm doing wrong and how to fix it.

Thanks!
 
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Hi EcKoh, welcome to PF.
In the relevant equations, the first equation is wrong. Check that.
 
Yes, vf != v0 + ad, rather, v^2 = v0^2 +2ad
 
What you are doing wrong is treat this two dimensional problem as if it were one-dimensional. You need to write the vertical and horizontal equations of motion. Then you need to consider at what part of the trajectory the speed is a minimum. Also, your equation
vf=vi+adv is incorrect.
 
Thanks guys, I corrected the formula and was able to get a correct answer of 6.69 m/s
 

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