A basketball player's hang time is the time spent in the air when shooting a basket. The formula [tex]t= \frac{ \sqrt{d} }{2} [/tex] models hang time, t, in seconds, in terms of the vertical distance of a player's jump, d, in feet. When a particular player dunked a basketball, his hang time for the shot was approximately 1.09 seconds. What was the vertical distance, d, of his jump, rounded to the nearest tenth?

As with any problem involving formulas, fill in the information you have and solve for the remaining variable.
.. 1.09 = [tex] \frac{ \sqrt{d}}{2} [/tex]
.. [tex] (2 \times 1.09)^{2} = d [/tex]
.. d ≈ 4.8 . . . . feet

Answer Link

Tringa0 Tringa0 · Answer 2 · 2019-10-29T10:04:20+01:00

Answer:

4.8 feet is the vertical distance.

Step-by-step explanation:

Basketball player's hang time(t) relationship with vertical distance(d) of a player's jump is given by:

[tex]t=\frac{\sqrt{d}}{2}[/tex]

[tex]2\times t=\sqrt{d}[/tex]

[tex]4t^2=d[/tex]

[tex]d=4t^2[/tex]

So, the vertical distance of player when hang time is equal to 1.09 seconds.

t = 1.09 s

[tex]d=4\times (1.09 s)^2[/tex]

d = 4.7524 feet ≈ 4.8 feet

4.8 feet is the vertical distance.