In basketball, hang is an illusion in which a player seems to weaken the gravitational acceleration while in mid-air. The illusion depends much on a skilled player’s ability to rapidly shift the ball between hands during the flight, but it might also be supported by the longer horizontal distance the player travels in the upper part of the jump than in the lower part. If a player jumps with an initial speed of${\mathbf{v}}_{\mathbf{o}}\mathbf{=}\mathbf{7}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{m}\mathbf{/}\mathbf{s}$at an angle of ${\mathbf{\theta }}_{\mathbf{o}}\mathbf{=}\mathbf{35}\mathbf{.}\mathbf{0}\mathbf{°}$, what percent of the jump’s range does the player spend in the upper half of the jump (between maximum height and half maximum height)?

It is given that,

$$\begin{gathered} {\mathrm{v}}_{\mathrm{o}}=7\mathrm{m}/\mathrm{s} \\ \mathrm{\theta }=35.0° \end{gathered}$$

The velocity of a player will be zero at maximum height. And, the initial velocity and projection angle of player are given. Now, find the time to reach the maximum height to player. And from this time, obtain quadratic equation for time.

After solving that quadratic equation, 2 values for time will get. Among which, one will be for ascending and other will be for descending at crossing of upper half level. Thus, the percentage difference between jump’s ranges can be obtained.

Formula:

$$\begin{gathered} {\mathrm{v}}_{\mathrm{f}}={\mathrm{v}}_{\mathrm{o}}+\mathrm{at} \\ \mathrm{s}={\mathrm{v}}_{\mathrm{o}}\mathrm{t}+\frac{1}{2}{\mathrm{gt}}^2 \end{gathered}$$

Using the fact that object has zero velocity at its maximum height, find the time for reaching the maximum height.

Let’s take the magnitude of an acceleration to find the position of object from axis.

$$\begin{gathered} \mathrm{v}={\mathrm{v}}_{\mathrm{o}}+\mathrm{at} \\ \mathrm{o}={\mathrm{\theta }}_{\mathrm{o}}\sin \mathrm{at} \\ \mathrm{t}=\frac{{\mathrm{\theta }}_{\mathrm{osin}}}{\mathrm{g}} \\ {\mathrm{t}}_{\mathrm{maximum}}=\frac{{\mathrm{\theta }}_{0\sin }}{\mathrm{g}} \end{gathered}$$

Now, putting above value in equation of maximum height,

$$\begin{gathered} {\mathrm{y}}_{\mathrm{maximum}}={\mathrm{v}}_0\left(\mathrm{sin\theta }\right){\mathrm{t}}_{\mathrm{maximum}}-\frac{1}{2}{{\mathrm{gt}}^2}_{\mathrm{maximum}} \\ {\mathrm{y}}_{\mathrm{maximum}}={\mathrm{v}}_0\left(\mathrm{sin\theta }\right)\frac{{\mathrm{v}}_0\mathrm{sin\theta }}{\mathrm{g}}-\frac{1}{2}\mathrm{g}\left(\frac{{\mathrm{v}}_0\mathrm{sin\theta }}{\mathrm{g}}\right) \\ {\mathrm{y}}_{\mathrm{maximum}}=\frac{{\mathrm{v}}_0^2{\mathrm{sin\theta }}^2}{2\mathrm{g}} \end{gathered}$$

Now, find the time at half of maximum height,

$$\begin{gathered} \mathrm{y}=\frac{{\mathrm{y}}_{\mathrm{maximum}}}{2}=\frac{{\mathrm{v}}_0^2{\mathrm{sin\theta }}^2}{4\mathrm{g}}=\left({\mathrm{v}}_0\mathrm{sin\theta }\right)\mathrm{t}-\frac{1}{2}{\mathrm{gt}}^2 \\ \frac{1}{2}{\mathrm{gt}}^2-\left({\mathrm{v}}_0\mathrm{sin\theta }\right)\mathrm{t}+\frac{{\mathrm{v}}_0^2{\mathrm{sin\theta }}^2}{4\mathrm{g}}=0 \\ \mathrm{t}\pm =\frac{\left(2\pm \sqrt{2}\right){\mathrm{v}}_0{\mathrm{sin\theta }}_0}{2\mathrm{g}} \end{gathered}$$

This is the time which player spends during crossing to half of maximum height.

Now, subtract this time from maximum height time. So, find the difference between them

$$\begin{gathered} △\mathrm{t}={\mathrm{t}}_{\mathrm{maximum}}-\mathrm{t} \\ △\mathrm{t}=\frac{{\mathrm{v}}_0\mathrm{sin\theta }}{\mathrm{g}}-\frac{\left(2\pm \sqrt{2}\right){\mathrm{v}}_0{\mathrm{sin\theta }}_0}{2\mathrm{g}} \\ △\mathrm{t}=\frac{{\mathrm{v}}_0\mathrm{sin\theta }}{\sqrt{2g}}=\frac{{\mathrm{t}}_{\mathrm{maximum}}}{\sqrt{2}} \end{gathered}$$

If ratio is taken of half maximum time and maximum time, the percentages of jump’s range does player spend in upper half of jump can be found,

$\frac{△\mathrm{t}}{{\mathrm{t}}_{\mathrm{maximum}}}=\frac{1}{\sqrt{2}}=0.707$

So, player spends 70.7% of time in upper half of the jump.