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

AI Thread Summary
The discussion focuses on a physics problem involving two balls thrown from a balcony, requiring calculations for the time spent in the air, their velocities upon impact, and their distance apart after a specific time. The original poster struggles with understanding the equations of motion and how to apply them to solve the problem. Key equations mentioned include the kinematic equations for vertical motion, specifically relating to displacement, velocity, and acceleration due to gravity. Contributors provide guidance on how to set up the equations correctly and emphasize the importance of solving for the time each ball spends in the air. The poster expresses gratitude for the assistance received in clarifying the concepts.
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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<br /> \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*}<br />
 

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here is a hint, use the equation: d = v_{o}t + \frac{1}{2}at^2. 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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