What is the difference in time the balls spend in the air?

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Homework Help Overview

The problem involves two students throwing balls from a balcony 19.6 meters above the ground, one ball thrown downward and the other upward, both at 14.7 m/s. The main question is to determine the difference in time the balls spend in the air.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to apply kinematic equations but expresses confusion about the setup, particularly regarding the second ball's height. Some participants suggest using specific equations of motion to analyze the problem.

Discussion Status

Participants are exploring various approaches to the problem, with some providing hints and equations. The original poster acknowledges the help received and expresses gratitude, indicating a supportive atmosphere in the discussion.

Contextual Notes

The original poster mentions having multiple similar problems, suggesting a broader context of homework constraints. There is also an indication of uncertainty regarding the second ball's trajectory and its relation to the balcony.

holly111
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I have been working on this alllll night PLEASE SOMEONE HELP! I never ask for help with this stuff but i really need help with this. I don't understand how to do these problems and i have 5 others. They all go something like this:


Two students are on a balcony 19.6 meters above the street. one student throws a ball vertically downward at 14.7 m/s. At the same instant, the other student throws a ball vertically upward at the same speed. The second ball just misses the balcony on the way down.


a. What is the difference in time the balls spend in the air?

b. What is the velocity of each ball as it strikes the ground?

c. How far apart are the balls 0.800s after they are thrown?

I've done all this:

s = d t = d d = s
- -
t s


a i t = d = 19.6
- -----
s 14.7

ii = 19.6 + 2x
---------
14.7


i don't understand how high the second ball i just don't know how to get the (x)

please help with this, i know I'm doing it all totally wrong.

Hollyxxx
 
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[tex] \begin{align*}<br /> Part\ (a)\\<br /> s=ut+\frac{1}{2}at^2\\<br /> y_1=-14.7t_1 -\frac{g}{2}t_1^2\ \ \ ...(1) \ where\ g=9.81\ m/s^2\ and\ t_1\ is\ the\ time\ ball\ 1\ spends\ in\ the\ air \ before\ hitting\ the\ street\\<br /> y_2=14.7t_2 -\frac{g}{2}t_2^2\ ...(2)\ where\ t_2\ is\ the\ time\ ball\ 2\ spends\ in\ the\ air\ before\ hitting\ the\ street\\<br /> y_1=y_2=-19.6\ m\\<br /> Solve\ for\ (2)-(1).\\ \\<br /> Part\ (b) \\<br /> Use \ v^2=u^2+2as\\ \\<br /> Part\ (c)\\<br /> Use\ equation\ (1)\ and\ (2)\ but\ use\ common\ t\ now\ which\ means\ for\ any\ given\ t,\ what\ are\ their\ positions\ relative\ to\ the\ balcony.\\<br /> Use\ |y_2-y_1|;\ the\ modulus\ sign\ gives\ the\ distance\ which\ is\ a\ scalar.<br /> \end{align*}[/tex]
 

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here is a hint, use the equation: [tex]d = v_{o}t + \frac{1}{2}at^2[/tex]. and follow what Leong did.
 
Thank you so much

I want to thank both Leong and Nenad!

Thank you both so much i apresiate your help greatly!

*hugs to you both*

:D Holly :D
 

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