Question

The time a basketball player spends in the air when shooting a basket is called the​ "hang time." the vertical leap l measured in feet is related to the hang time t measured in seconds by the equation L=4t2. suppose that a basketball player has a vertical leap of 2 feet 2 inches. find the hang time for this leap

Answer 1

The hang time for this leap is 0.736 seconds

Solution:

The vertical leap l measured in feet is related to the hang time t measured in seconds by the equation:

$L = 4t^2$

Where,

L is the vertical leap in feet

t is the hang time in seconds

From given,

L = 2 feet 2 inches

$1\ inch = \frac{1}{12}\ feet\\\\2\ inches = \frac{2}{12} = 0.1667\ feet$

Therefore,

L = 2 feet + 0.1667 feet = 2.1667 feet

Substituting in equation,

$2.1667 = 4 \times t^2\\\\t^2 = \frac{2.1667}{4}\\\\t^2 = 0.541675\\\\t = 0.73598 \approx 0.736$

Thus the hang time for this leap is 0.736 seconds