Determine the flight time and heights of 2 balls thrown vertically

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Discussion Overview

The discussion revolves around determining the flight time and heights of two balls thrown vertically. Participants explore the application of kinematic equations and projectile motion principles to solve the problem, focusing on both time and height calculations.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant calculated the time using the equation 3 + (10 * x) = 13.5 * x, arriving at x = 0.857, but struggled to find the height.
  • Another participant requested to see the work done to find the height.
  • A different participant proposed using the projectile motion formula y = -4.9t^2 + vo*t*sinO, but noted it did not lead to the correct height.
  • One participant questioned the relevance of an angle in the context of vertical motion and suggested the original poster (OP) review basic projectile motion concepts.
  • Another participant acknowledged the OP's time calculation as correct but noted the need to replace "vo" with 13.5 m/s and clarified that the angle of projection is 90 degrees for vertical throws. They also provided a corrected height calculation of 7.971 m.

Areas of Agreement / Disagreement

There is no consensus on the correct approach to finding the height, as participants have differing views on the relevance of projectile motion equations and the calculations involved. Some agree on the correctness of the time calculation, while others challenge the methods used.

Contextual Notes

Participants have not fully resolved the assumptions regarding the use of projectile motion equations, and there are unresolved steps in the calculations for height. The discussion reflects varying levels of understanding of the underlying physics concepts.

tremain74
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Homework Statement
Child A throws a ball vertically up with a speed of 10 m/s from the top of a shed 3 meters high. Child B on the ground at the same instant throws a ball vertically up with a speed of 13.5 m/s. Determine the time at which the two balls will be at the same height above the ground. What is the height? The answers were t = 0.857 s and h = 15.17.
Relevant Equations
I used the equations 3 + (10*x) = 13.5*x.
To find the time of this problem, I used the equations 3 + (10 * x ) = 13.5 *x. 3 = 3.5*x. Therefore x was 0.857. To start off with child A was already three feet above ground and I got the answer to the time. However I couldn't find the height.
 
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show what work you did do to try to get the height.
 
The one formula that I thought about was the projectile motion formula using y = -4.9t^2 + vo*t*sinO. Sin O(sin of theta). However this does not solve my problem to find the correct height.
 
Since the balls were thrown vertically, why would you think there would be an angle involved. Do you have ANY experience with projectile motion? It certainly seems as though you do not, so I would suggest you go back and study the basics of projectile motion.
 
Surprisingly the OP's answer for the first part is correct albeit using variable x instead of t. This is because in problems like this the quadratic term(s) in the respective motion equations cancel out and one is left with exactly the equation solved by the OP.

The OP also wrote the correct equation to find the height - just needed to replace "vo" with 13.5 m/s. and the angle of projection is 90 degrees if the ball is thrown vertically upward.

The given answer for the height at which the balls meet (are in same horizontal line) seems incorrect - should be 13.5*0.857-4.9*(0.857)^2=7.971 m. See following graph.

https://www.desmos.com/calculator/snmbofcveh
 
Last edited:

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