What is the speed at which the object was thrown upward?

AI Thread Summary
The discussion revolves around calculating the speed at which an object is thrown upward when it passes another object dropped from a height H at H/2. The user employs kinematic equations, setting gravitational acceleration as g = -10 m/s², and establishes two equations for the motion of both objects. By equating their heights at the moment they pass, the user derives the initial speed of the thrown object as vbi = sqrt(H * abs(g)). The calculations involve determining the time t1 it takes for both objects to reach H/2, leading to the final expression for speed. The solution demonstrates a solid understanding of basic physics principles despite initial uncertainty.
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Homework Statement


You throw an object directly upward at the same time a person drops an object down to you from a height, H. The two objects pass at a height H/2. With what speed did you throw the object upward?


Homework Equations


Simple kinematics equations.


The Attempt at a Solution


I haven't done physics for a while so I'm not certain if I'm entirely off track on this one:

I set up two equations for the motion of the balls (I'm allowing g = -10m/s/s); hence, I got h_a = (gt^2)/2 + H and h_b = (gt^2)/2 + vbi(t) where h_a is the height of the ball being dropped and h_b is the height of the ball being thrown (sorry if my notation is poor or non-standard). Since the balls are released simultaneously and pass at height H/2, I allow t1 to designate the amount of time required for the balls to reach H/2. By equating the two (h_a = h_b = H/2) I get (gt1^2)/2 + H = (gt1^2)/2 + vbi(t1), which solving for vbi produces vbi = H/t1. To solve for t1 I set up the equation H/2 = (gt1^2)/2 + H, and with some manipulations get t1^2 = -H/g and consequently t1 = sqrt(H/abs(g)). Using my value for t1 in my equation for vbi yields vbi = H(sqrt(abs(g)/H) = sqrt(Habs(g)).

Sorry the work is so messy, I don't know how to use Latex.
 
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Looks good to me! Nicely done.
 
Thanks!
 
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