IF HANG TIME FOR A SHOT BY A PROFESSIONAL BASKETBALL PLAYER IS O.85 SECONDS WHAT IS THE VERTICAL DISTANCE OF THE JUMP ROUNDED TO THE NEAREST TENTH OF A FOOT?

falls for .85/2 = .425 second

h = (1/2) g t^2 = (1/2)(32)(.425^2)
= 2.89 ft = 2.9 ft

To determine the vertical distance of the jump, we can use the equation of motion for vertical motion, assuming the basketball player jumps vertically without any horizontal motion. The equation is:

h = (1/2) * g * t^2

Where:
h = vertical distance
g = acceleration due to gravity (approximately 32.2 ft/s^2)
t = time the player is in the air (hang time)

Let's calculate the vertical distance:

h = (1/2) * 32.2 ft/s^2 * (0.85 seconds)^2
h = 13.7457 ft

Rounded to the nearest tenth:
h ≈ 13.7 ft

Therefore, the vertical distance of the jump, rounded to the nearest tenth of a foot, is approximately 13.7 feet.