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 $t = 2(\frac{2h}{32})^{\frac{1}{2}}$, 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.

Explanation:

Step1: Solve for $h$ in the hang - time formula

Given $t = 2(\frac{2h}{32})^{\frac{1}{2}}$, first simplify the formula: $t = 2(\frac{h}{16})^{\frac{1}{2}}=\frac{2}{\sqrt{16}}\sqrt{h}=\frac{1}{2}\sqrt{h}$. Then, solve for $h$: $\sqrt{h}=2t$, so $h = 4t^{2}$.

Step2: Calculate the height of Player 1

For Player 1 with $t_1 = 0.9$s, substitute $t_1$ into the formula $h = 4t^{2}$. So $h_1=4\times(0.9)^{2}=4\times0.81 = 3.24$ feet.

Step3: Calculate the height of Player 2

For Player 2 with $t_2 = 0.8$s, substitute $t_2$ into the formula $h = 4t^{2}$. So $h_2=4\times(0.8)^{2}=4\times0.64 = 2.56$ feet.

Step4: Find the difference in heights in feet

The difference in heights $\Delta h=h_1 - h_2=3.24 - 2.56 = 0.68$ feet.

Step5: Convert the difference to inches

Since 1 foot = 12 inches, $\Delta h_{inches}=0.68\times12 = 8.16\approx8$ inches.

Answer:

8