Question

In basketball, hang time is the time that both of your feet are off the ground during a jump. The equation for hang time is t=2(2h/32)^1/2, where t is the time in seconds, and h is the height of the jump, in feet.

Player 1 had a hang time of 0.9 s. Player 2 had a hang time of 0.8 s. To the nearest inch, how much higher did Player 1 jump than Player 2?
____in.
is it 8? because I solved the problem and player 1s jump height minus player 2 is 0.68 feet which is 8.16 inches?

Answer 1

hi, that answer is right BUT i just want to let everyone know that you should use an app call PHOTO MATH it helps so much, cuz it takes to long to type everything in

Answer 2

We have an equation:
t = 2 ( 2h/32 ) ^1/2 ( or: t = 2 * sqrt ( 2h/32) ).
For the 1st player:
0.9 = 2 * sqrt( 2h/32 ) / ^2 ( we will square both sides of equation )
0.81 = 4* 2h/32
0.81 = h/4
h 1 = 0.81 * 4 = 3.24 ft
For the 2nd player:
0.8 = 2 * sqrt( 2h/32 ) /^2
0.64 = 4 * 2 h/32
h 2 = 0.64 * 4 = 2.56 ft
Finally : h 2 - h 1 = 3.24 - 2.56 = 0.68 ft
and since 12 in = 1 ft:
0.68 * 12 = 8.16 in ≈ 8 in.
Answer: 8 in.