Kinematics in one dimention, where do two balls meet?

[feihong47]

1. You drop a ball from a window located on an upper floor of a building. It strikes the ground with speed v. You now repeat the drop, but your friend down on the ground throws another ball upward at the same speed v, releasing her ball at the same moment that you drop yours from the window. At some location, the balls pass each other. Is this location (a) at the halfway point between window and ground, (b) above this point, or (c) below this point?

Someone said the solution is b. Subjectively this may be find and dandy, but does any have proof of this using equations? I tried to find the time of each ball to reach halfway and then compare. Using the dropped ball as t1, i get t1 = SQRT(h/g), where h is the height of the building. Using t2 as the thrown ball, i have to solve t2 as a quadratic formula (1/2 h = vt-1/2gt2), in which case i have t2 in terms of v, g, and h. however I had no way to compare the two time intervals.

 

[wukunlin]

you want to find the intersection of two displacement vs time curves
the first one is the ball dropping from the upper floor:
d = - \frac{1}{2}gt^2

the other one is the ball being tossed up
d = vt - \frac{1}{2}gt^2 - h
where h is the height between the floors

can you see how to solve it now?

 

[feihong47]

i set the two equations equal to each other and i get h=vt. not sure how to use that, so i solve for t and plug it back in EQ 1 and i get d = -4.9h²/v² ... doesn't really help yet

another thing is -- for ball being tossed up, i set up the equation as d=vt−1/2gt2 but without - h... how did you set yours up like that to make sense?

 

[vela]

How are v and h related?

 

[wukunlin]

about the -h:

the expressions for d need to have the same origin, in the equations I have written, d=0 is the release point at the upper floor, therefore the equation for the ball being tossed up, d must be equal to -h at time t=0.

if you like you can define the d=0 at any other point, your answer won't change qualitatively

 

[feihong47]

you want to find the intersection of two displacement vs time curves
the first one is the ball dropping from the upper floor:
d = - \frac{1}{2}gt^2

the other one is the ball being tossed up
d = vt - \frac{1}{2}gt^2 - h
where h is the height between the floors

can you see how to solve it now?

ok, for some reason I don't see it..

i understand the intersection of the two curves is the point where the two balls meet.

therefore I set the two equations equal to each other and get :

vt = h

I solved for t = h/v, and then substituting into EQ 1, and I get

d = - \frac{1}{2}g(\frac{h}{v})^2

but how does that indicate where exactly the balls are?

 

[wukunlin]

what is v in terms of h?

 

[feihong47]

what is v in terms of h?

thanks! got it now so d = -h/4, so that indicates it's above the halfway line (rather than the bottome 1/4) right?

 

[wukunlin]

thanks! got it now so d = -h/4, so that indicates it's above the halfway line (rather than the bottome 1/4) right?

precisely :)