Question

A basketball player grabbing a rebound jumps 76.0 cm vertically. How much total time (ascent and descent) does the player spend (a) in the top 15.0 cm of this jump and (b) in the bottom 15.0 cm? Do your results explain why such players seem to hang in the air at the top of a jump?

Answer 1

a) we can answer the first part of this by recognizing the player rises 0.76m, reaches the apex of motion, and then falls back to the ground we can ask how

long it takes to fall 0.13 m from rest: dist = 1/2 gt^2 or t=sqrt[2d/g] t=0.175

s this is the time to fall from the top; it would take the same time to travel

upward the final 0.13 m, so the total time spent in the upper 0.15 m is 2x0.175

= 0.35s

b) there are a couple of ways of finding thetime it takes to travel the bottom 0.13m first way: we can use d=1/2gt^2 twice

to solve this problem the time it takes to fall the final 0.13 m is: time it

takes to fall 0.76 m - time it takes to fall 0.63 m t = sqrt[2d/g] = 0.399 s to

fall 0.76 m, and this equation yields it takes 0.359 s to fall 0.63 m, so it

takes 0.04 s to fall the final 0.13 m. The total time spent in the lower 0.13 m

is then twice this, or 0.08s

Answer 2

Answer:

Part a)

$T = 0.35 s$

Part b)

$T' = 0.041 s$

So for top 15 cm the time interval is sufficiently larger than the time interval of last 15 cm

Explanation:

Part a)

Maximum height reached is

H = 76 cm

now the velocity of the player at starting position is given as

$v_f^2 - v_i^2 = 2ad$

$0 - v^2 = 2(-9.81)(0.76)$

$v = 3.86 m/s$

time taken by it in top 15 cm position is given as

$y = \frac{1}{2}gt^2$

$0.15 = \frac{1}{2}(9.81)t^2$

$t = 0.175 s$

so total time in that position is double because in that position first it will go up and then go down

$T = 0.35 s$

Part b)

Now for bottom position of 15 cm first we will find the time to reach (76 - 15)cm have

$H = \frac{1}{2}gt^2$

$(0.76 - 0.15) = \frac{1}{2}(9.81)t^2$

$t_1 = 0.352 s$

now for total time to drop

$t_2 = \sqrt{\frac{2H}{g}}$

$t_2 = 0.394 s$

so time interval of last 15 cm is given as

$t' = 0.394 - 0.352$

$T' = 0.041 s$

So for top 15 cm the time interval is sufficiently larger than the time interval of last 15 cm