Question

A basketball​ player's hang time is the time spent in the air when shooting a basket. The formula
`t= ( √(d) )/(2)` 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

4.8 feet is the vertical​ distance.

Explanation:

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

`t=(√(d))/(2)`

`2* t=√(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* (1.09 s)^2`

d = 4.7524 feet ≈ 4.8 feet

4.8 feet is the vertical​ distance.

Answer 2

As with any problem involving formulas, fill in the information you have and solve for the remaining variable.
.. 1.09 =
`( √(d))/(2)`
..
`(2 * 1.09)^(2) = d`
.. d ≈ 4.8 . . . . feet