hang time in seconds 1

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 L4t2.
Suppose that a basketball player has a vertical leap of 4
feet 4
inches. Find the hang time for this leap.

Answer 1

Hi Jessica,

 

I am not clear regarding your formula. I don't know if you have taken Physics, but bear with me. Let me get the right formula for hang time. You can probably ignore the Physics calculation, and I can tell you when to refocus when we get to the Algebra.

 

When the basketball player jumps up, he has an initial velocity v₀ but his final velocity at the pinnacle of his leap is v = 0. We are told that his maximum height of his leap is L. We can use a kinematic equation from Physics to calculate his leap as a function of his initial vertical velocity and the acceleration of gravity g.

 

v² = v₀² - 2gL

 

But the final velocity v = 0 and so we can write

 

L = v₀²/2g

 

We also know another kinematic equation from Physics involving time:

 

v = v₀ -gt

 

Again, we know that the final velocity v = 0, so we can express the initial velocity as

 

v₀ = gt

 

We can substitute for v₀ in the first equation for L with this latest expression and write

 

L = (gt)² / 2g = g²t² / 2g = gt²/2 canceling out a g.

 

t = √[2L/g]

 

But, this time t is just the time for the basketball player to ascend. The hang time is his time in the air, so his time of ascension and descent. Assuming symmetry, the hang time is twice the time of ascension, or hang time t_h = 2t

 

t_h = 2√[2L/g]

 

We are told that the leap for a given basketball player is  L = 4 feet 4 inches = 4.33 feet since there are 12 inches in a foot.

 

g = 32 feet/sec².

 

so, t_h = 2√[2(4.33)/(32)] = 2√[0.2706] = 1.04 sec

 

Where did you get your formula?