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Mathias
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Homework Statement
Suppose you drop a ball from rest out of a window that is at height H above the ground, and suppose the ball hits the ground with speed S.
At the instant that you release the ball, your buddy throws a ball straight up from the ground level with the same speed S.
Show that the balls pass each other at a height of 3/4H
What we know: for person dropping the ball
##y_o = h##
##y_1=h##
##a_y=-9.81m/s^2##
##v_0=0m/s##
##v_y = S##
Person launching the ball up:
##y_o = 0##
##y_1= h##
##a_y=-9.81m/s^2##
##v_0= S##
##v_y = 0m/s##
Homework Equations
##v_y = v_0 + a_yt##
##y=y_0+v_0t+1/2a_yt^2##
##v^2_y = v^2_0+2a_y(y-y_0)##
##y-y_0=1/2(v_0+v_x)t##
## v^2_y-v^2_0=2a_y(y-y_0) ##
The Attempt at a Solution
So my thought process is I need to use a kinematic equation to solve for a value let's say t, for both people, one dropping the ball and one throwing the ball.
Eq.4 ## y-y_0=1/2(v_0+v_x)t ## -> ## t=2(y-y_0)/(v_0+v_y) ##
so now that we have an equation for a function of time, my thought process is that the time when the person throws the ball the height should be 3/4H. I'm not sure if I can do this, but I'm hoping I can set v_y = 0, I think I can use it for the person dropping the ball, but not for the person throwing the ball, but I got stuck here if I didn't.
##t=2(3/4H)/(S+0) ## -> t = 3H/2S
Then for the person throwing the ball down
## t= 1/2(1/4H)/0+S ## -> t=H/2S
then when you set them equal to each other you get H/2s = 3H/2s -> 3
I'm hoping in some way shape or form that is still a t, then to test my answer I plug t=3 into equation 4 again.
##dropping: y-y_0= 1/2(v_0+v_y)t## -> ## 1/4H= 1/2(0+S)3 ## -> ##H=6S##
##throwing: y-y_0 = 1/2(v_0+v_y)t ## -> ## 3/4H=1/2(S+0)3 ## -> ##H=2S##
Set them equal to each other to get 3 again. I think at this point its a huge stretch, and it's probably just an algebra reason why I got 3 again. I'm not sure how else to approach this.