A basketball player grabbing a rebound jumps 80 cm vertically. How much total time (ascent and descent) does the player spend (a) in the top 10 cm of this jump and (b) in the bottom 10 cm?

Answer 1

For constant acceleration, the height as a function of time is

y(t) = v₀t – gt²/2

that returns to y(t) → 0 at

t → 2v₀/g = T.

So y(t) = (T – t)gt/2

and has a maximum y → Y = 0.8m at the root of

y'(t) = gT/2 – gt,

namely at t → T/2.

So, Y = (T – T/2)g(T/2)/2 = gT²/8 = v₀²/2g.

Therefore y(t) = 4(T – t)tY/T²

is manifestly symmetric about t → T/2 = √(2Y/g).

For t < T/2 we can invert y(t) as

0 = t² – Tt + T²y/4Y = (t – T/2)² – (1 – y/Y)T²/4 ⇒ t(y) = (1 – √(1 – y/Y))T/2.

To initially reach y → Y/8 = 0.1m takes t(Y/8) = (1 – √(7/8))T/2, so by symmetry, 2t(Y/8) is the time spent below Y/8. To pass from y → 7Y/8 to Y takes τ = t(Y) – t(7Y/8) = (1/√2)T/4, so by symmetry, 2τ is the time spent above 7Y/8.