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

[holly111]

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

 

[Leong]

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

 

[Nenad]

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

 

[holly111]

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