Answered

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 [tex]t=2\left(\frac{2 h}{32}\right)^{\frac{1}{2}}[/tex], where [tex]t[/tex] is the time in seconds, and [tex]h[/tex] 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?

[tex]\quad[/tex] inches.



Answer :

GinnyAnswer
To solve the problem of finding the difference in jump height between Player 1 and Player 2, we need to use the hang-time formula and perform several calculations. Let's go through the steps:

1. Understand the hang-time equation:
[tex]\[ t = 2 \left( \frac{2h}{32} \right)^{\frac{1}{2}} \][/tex]

2. Rearrange the equation to solve for height (h):
[tex]\[ t/2 = \left( \frac{2h}{32} \right)^{\frac{1}{2}} \][/tex]
Squaring both sides:
[tex]\[ \left( \frac{t}{2} \right)^2 = \frac{2h}{32} \][/tex]
Solving for [tex]\( h \)[/tex]:
[tex]\[ h = \left( \frac{t}{2} \right)^2 \cdot \frac{32}{2} \][/tex]
[tex]\[ h = \left( \frac{t}{2} \right)^2 \cdot 16 \][/tex]

3. Calculate the height for Player 1:
- Player 1's hang time, [tex]\( t_1 = 0.9 \)[/tex] seconds
[tex]\[ h_1 = \left( \frac{0.9}{2} \right)^2 \cdot 16 \][/tex]
[tex]\[ h_1 = (0.45)^2 \cdot 16 \][/tex]
[tex]\[ h_1 = 0.2025 \cdot 16 \][/tex]
[tex]\[ h_1 = 3.24 \text{ feet} \][/tex]

4. Calculate the height for Player 2:
- Player 2's hang time, [tex]\( t_2 = 0.8 \)[/tex] seconds
[tex]\[ h_2 = \left( \frac{0.8}{2} \right)^2 \cdot 16 \][/tex]
[tex]\[ h_2 = (0.40)^2 \cdot 16 \][/tex]
[tex]\[ h_2 = 0.16 \cdot 16 \][/tex]
[tex]\[ h_2 = 2.56 \text{ feet} \][/tex]

5. Find the difference in heights:
[tex]\[ \Delta h = h_1 - h_2 \][/tex]
[tex]\[ \Delta h = 3.24 - 2.56 \][/tex]
[tex]\[ \Delta h = 0.68 \text{ feet} \][/tex]

6. Convert the height difference from feet to inches:
[tex]\[ 1 \text{ foot} = 12 \text{ inches} \][/tex]
[tex]\[ \Delta h_{\text{inches}} = 0.68 \times 12 \][/tex]
[tex]\[ \Delta h_{\text{inches}} = 8.16 \text{ inches} \][/tex]

7. Round the result to the nearest inch:
[tex]\[ \Delta h_{\text{inches}} \approx 8 \text{ inches} \][/tex]

Therefore, Player 1 jumped approximately 8 inches higher than Player 2.

Other Questions