Solving the Unsolvable: Two Ball Problem

[saiyajin822]

ive been having difficulty with this problem also:

A ball is thrown downward with an initial speed of 25m/s from the top of a 210 m tall building. At the same time, another ball is thrown upward from ground level with a speed of 25m/s. At what distance from the bottom of the two balls pass each other?

the answer is supposed to be "-10 m so it is illogical."
so far i know that for the first ball Vi=25m/s d=210m a=9.8m/s^2

2nd ball Vi=25n/s d=? a=-9.8m/s^2

are we supposed to set the distance of each of the balls equal?? i really have no idea how to do this

 

[Tide]

Write the position as a function of time for each of the balls then set them equal to find the time at which they meet.

I got 18.56 m.

 

[Pyrrhus]

Find a equation for each ball in function of distance and equal them.Look up Kinematic Equations for Uniform Acceleration, here the acceleration is Gravity (might want to look up Free Fall, too).

 

[saiyajin822]

so the equation would be (25m/s)(t)+1/2(9.8)t^2=(25m/s)(t)+1/2(-9.8)t^2?
(1st ball and 2nd ball respectively)
what do i do with the 210 m?

btw, my teacher gave us an answer sheet and the answer is -10m

 

[Pyrrhus]

Saiyan... Stick to your Sign Convention... if put down and left is negative and up and right is positive, then this should affect the systems you're studying, i see a positive acceleration for one and a negative for another.

 

[Pyrrhus]

I also noticed improper use of the equation.

$$Y - Y_{o} = V_{o}t + \frac{1}{2}at^2$$

where, Yo, Vo and a is known.

 

[saiyajin822]

that is the equation i used: d=Vi(t)+1/2a(t)^2

 

[Pyrrhus]

that is the equation i used: d=Vi(t)+1/2a(t)^2

For this problem you'll have to use it with Y - Yo.

 

[saiyajin822]

Y-Yo? i only see one y and that's 210m