Question

A basketball​ player's hang time is the time spent in the air when shooting a basket. The formula /?f=t%3D%20%5Cfrac%7B%20%5Csqrt%7Bd%7D%20%7D%7B2%7D%20" id="TexFormula1" title="t= \frac{ \sqrt{d} }{2} " alt="t= \frac{ \sqrt{d} }{2} " align="absmiddle" class="latex-formula"> 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?

Answer 1

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

Answer 2

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:

$t=\frac{\sqrt{d}}{2}$

$2\times t=\sqrt{d}$

$4t^2=d$

$d=4t^2$

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

t = 1.09 s

$d=4\times (1.09 s)^2$

d = 4.7524 feet ≈ 4.8 feet

4.8 feet is the vertical​ distance.