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

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SUMMARY

The problem involves calculating the initial speed at which an object is thrown upward while another object is dropped from a height H. The two objects meet at a height of H/2. By applying kinematic equations and setting the gravitational acceleration g to -10 m/s², the solution derives the initial speed (vbi) of the thrown object as vbi = sqrt(H * abs(g)). This conclusion is reached by equating the heights of both objects at the midpoint and solving for time (t1) and initial velocity.

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jgens
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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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