Calculating Time Interval in Two-Dimensional Kinematics

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Homework Help Overview

The discussion revolves around a two-dimensional kinematics problem involving the calculation of time intervals between successive stroboscopic exposures of a ball launched from a cart. The problem includes a specific height measurement and requires estimating the time based on the vertical distance the ball travels.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the initial velocity of the ball and its implications for the calculations. Some question the accuracy of the initial velocity value used. Others explore different methods to calculate the time interval, including using equations of motion and considering the number of flashes in relation to the distance fallen.

Discussion Status

There are various interpretations of the problem setup and calculations. Some participants have provided alternative methods and values, while others have expressed uncertainty about the initial conditions and the number of intervals to consider. A participant noted a discrepancy in their calculations, leading to further exploration of the problem.

Contextual Notes

Participants mention the need for precision in calculations and the potential for misunderstanding the problem's requirements, such as the format of the answer. There is also a reference to the specific number of flashes and how they relate to the time intervals being calculated.

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Homework Statement


In the photograph , suppose the cart that launches the ball is 13 cm high. The vertical distance from the lowest ball to the highest is about two cart heights or 26 cm.

Estimate the time interval between successive stroboscopic exposures.

Link to the photograph: http://session.masteringphysics.com/problemAsset/1040728/5/Walker.4.44.jpg

Homework Equations


Y=y0+V0t+1/2at



The Attempt at a Solution


.26=0+2.3t+1/2*-9.81t^2
.26=2.3t-4.905t^2
-4.905t^2+2.3t-.26=0

t=-2.3+-sqrt 2.3^2-4*-4.905*-.26/-9.81
t=-2.3+-sqrt 5.29+19.62*-.26/-9.81
t=-2.3+-sqrt 5.29-5.1/-9.81
t=-2.3 +- sqrt .19/-9.81
t= 2.3+- .44/-9.81
t= .19, .28
t= .19/3= .06
 
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I'm not sure I understand what is happening in that diagram. I don't understand your initial velocity of 2.3.

Would it be correct to say the ball falls 26 cm from the turnover point in 3 flashes? If so you can find the answer quickly by using y = .5*g*t^2 with y and g both positive.
 
Well, I know the initial velocity is right, its a two part problem and that was the answer to part A, as for the diagram it's multiple pictures of a single ball being launched in an arc shape. I need to find the time between each picture. I tried finding the time to reach the max height, and dividing that by 3 (since there's three gaps between the first picture and that picture)
 
Okay, your work all looks good based on the 2.3 initial velocity. I got slightly different answers using
t = sqrt(2*.26/9.81) = .23 s for three flashes during the fall. I think the difference is due to your 2.3 being slightly inaccurate. If you use v^2 = 2ad = 2g*.26 you get v = 2.26 instead of 2.3.

The rule I'm used to for accuracy is to keep an extra digit in all calculations, rounding only the final answer when you report it.
 
mmkay, i tried that, and I also got .23. But .23/3 gave me .063... and the answer was still incorrect, am I dividing by the right number of time intervals?
 
Thanks for your help Delphi, but I figured it out myself, the computer just wanted it in scientific notation for some reason >_>
 

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