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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, H (called 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 t=0, and Boris jumps later, at time tR (his reaction time). Assume that Arabella has not yet reached her maximum height when Boris jumps.

Respuesta :

Preciousorekha1

Answer:

(a). D(t) = (√2gh)t - 1/2gt²

(b). D(t) = tk [(√2gh) + g(tk - 2t) / 2]

Explanation:

we would be using the equation of motion to determine the distance D

from the question;

  • the height of the hand raised by Boris is

H-boris = h₀ + V₀ (t - tk) - 1/2g(t -tk)²

  • Also given is the height of Arabella raised hand is given thus;

H-arabella = h₀ + V₀t - 1/2gt²

(a). the vertical displacement is given;

D(t) = H-arabella - H-boris

D(t) = h₀ + V₀t -1/2gt₂ - (h₀ + V₀(t-tk) -1/2g(t-tk)₂)

D(t) = h₀ + V₀t - 1/2gt₂ - h₀ where 0 ∠ t ∠ tk

this gives D(t) = V₀t -1/2gt²

where V₀ = √2gh

∴ D(t) = (√2gh)t - 1/2gt²

(b). We already know vertical displacement as;

D(t) = H-arabella - H-boris

D(t) = h₀ + V₀t -1/2gt₂ - (h₀ + V₀(t-tk) -1/2g(t-tk)₂)

= V₀tk - 1/2gt² + 1/2g(t -tk)²

  = V₀tk + 1/2gtk² - gttk

  = √2gh tk + 1/2gt² - gttk

this gives D(t) = tk [(√2gh) + g(tk - 2t) / 2]

cheers i hope this helps.

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