Jump shot in basketball...?

Question

Two basketball players are essentially equal in all respects. (They are the same height, they jump with the same initial velocity, etc.) In particular, by jumping they can raise their centers of mass the same vertical distance, (called H their "vertical leap"). The first player, Arabella, wishes to shoot over the second player, Boris, and for this she needs to be as high above Boris as possible. Arabella jumps at time , and Boris jumps later, at time (his reaction time). Assume that Arabella has not yet reached her maximum height when Boris jumps.



1)Find the vertical displacement , D(t) = H(a)(t) - H(b)(t) as a function of time for the interval 0 < t < t(r) , where H(a)(t) is the height of the raised hands of Arabella, while H(b)(t) is the height of the raised hands of Boris.
Express the vertical displacement in terms of H, g , and t.

2)Find the vertical displacement D(t) between the raised hands of the two players for the time period after Boris has jumped (t > tr) but before Arabella has landed.

Express your answer in terms of t, tr, g, and H.

______________________________

PLEASE TRY TO NOT GIVE ME AN OUTRIGHT ANSWER, BUT RATHER GIVE ME SOME GUIDANCE AS A PROFESSOR WOULD. Thanks for any help!

MISSED VARIABLES IN ORIGINAL QUESITON:

Arabella jumps at time t=0
Boris jumps at time t(r) (reaction time)

Answer 1

The notation will drive me nuts.
1) is asking for relative displacement when B hasn't jumped yet.
v0 = sqrt(2gH)
xa = v0t - gt^2/2 = sqrt(2gH)t - gt^2/2

2) After B has jumped and t>tr:
xa = t*sqrt(2gH) - g/2*t^2
xb = (t-tr)*sqrt(2gH) - g/2*(t-tr)^2
xa-xb = tr(sqrt(2gH)) - g/2 * (t^2-(t-tr)^2)
t^2-(t-tr)^2 = -tr^2+2t*tr = -tr(tr-2t)
Ans. deltax = xa-xb = tr(sqrt(2gH) + g(tr-2t)/2)

Answer 2

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