A basketball player, about to \"dunk\" the ball, jumps 62 cm vertically. (a) How

ID: 2000678 • Letter: A

Question

A basketball player, about to "dunk" the ball, jumps 62 cm vertically.

(a) How much (total) time does the player spend in the top 15 cm of this jump? (in ms)

(b) How much (total) time does the player spend in the bottom 15 cm of this jump? (in ms)

* No, it is an optical illusion and they spend the same amount of time at the tops and bottoms of their jumps.

* No, it is an optical illusion and they spend less time at the tops of their jumps.

* Yes, they spend more time at the tops of their jumps.

here,

height of jump, h = 62 cm

h = 0.62 m

for the top 15 cm of jump taking half the trip,

final velocity , v = 0

distance travelled , d = 0.15 m

using third equation of motion

v^2 - u^2 = - 2 * g * d

0 - u^2 = - 2 * 9.8 * 0.15

u = 1.71 m/s

let the time taken be t

v = u + g * t

0 = 1.71 - 9.8 * t

t = 0.175 s

the total time player spend in the top 15 cm is 2 * t = 2 * 0.175 = 0.35 s = 350 ms

for the bottom 15 cm of jump

distance travelled , d = 0.15 m

taking velocity at the bottom 15 cm be v

using third equation of motion

v^2 - u^2 = - 2 * g * (0.62 - d)

v^2 - 0^2 = - 2 * 9.8 * 0.47

u = 3.035 m/s

let the time taken be t

d = u * t + 0.5 * g * t^2

0.15 = 3.035 * t + 0.5 * 9.8 * t^2

t = 0.046 s

time taken in the bottom 15 cm is 2 * t = 2 * 0.046 = 0.092 s = 92 s

the time taken in the top 15 cm is higher than in the lower 15 cm

therefore, the player seem to hang in the air at the top of their jumps

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