We choose the origin to be located at the initial position of Tony Parker $\left(x_0=0,\ y_0=0\right)$. The initial velocity as he jumps has components $v_{0x}=v_0{\mathrm{cos} \left(\theta \right)\ }$ and $v_{0y}=v_0{\mathrm{sin} \left(\theta \right)\ }$ and his acceleration during his airborne motion has components $a_x=0$ and $a_x=-g$.

Thus, the kinematic equations for Tony Parker’s motion are given by:

\left\{ \begin{array}{c} \displaystyle{x\left(t\right)=x_0+v_{0x}t+\frac{1}{2}a_xt^2=v_0{\mathrm{cos} \left(\theta \right)\ }t} \\ \\ v_x\left(t\right)=v_{0x}+a_xt=v_0{\mathrm{cos} \left(\theta \right)\ } \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \end{array} \right. \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \left\{ \begin{array}{c} \displaystyle{y\left(t\right)=y_0+v_{0y}t+\frac{1}{2}a_yt^2=v_0{\mathrm{si}\mathrm{n} \left(\theta \right)\ }t-\frac{1}{2}gt^2} \\ \\ v_y\left(t\right)=v_{0y}+a_yt=v_0{\mathrm{sin} \left(\theta \right)\ }-gt \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \end{array} \right.

2. What maximum height $y_{max}$ does he reach? Where along the $x$-axis is this position located?

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He reaches his maximum height when his vertical velocity cancels. This happens for a time $t_1$ such that $v_y\left(t_1\right)=0$ and we solve

\begin{aligned} v_y\left(t_1\right)=0\ \ \ \ &\Leftrightarrow \ \ \ \ \ v_0{\mathrm{sin} \left(\theta \right)\ }-gt_1=0 \\ \\ &\Leftrightarrow \ \ \ \ \ t_1=\frac{v_0{\mathrm{sin} \left(\theta \right)\ }}{g} \end{aligned}

His height at time $t_1$ is his maximum height which is given by

y_{max}=y\left(t_1\right)=v_0{\mathrm{sin} \left(\theta \right)\ }t_1-\frac{1}{2}gt^2_1=\frac{v^2_0{{\mathrm{sin}}^2 \left(\theta \right)\ }}{g}-\frac{v^2_0{{\mathrm{sin}}^2 \left(\theta \right)\ }}{2g}=\frac{v^2_0{{\mathrm{sin}}^2 \left(\theta \right)\ }}{2g}

The maximum height reached by Tony Parker is equal to

\boxed{y_{max}=\frac{v^2_0{{\mathrm{sin}}^2 \left(\theta \right)\ }}{2g}=\frac{{10}^2\cdot {{\mathrm{sin}}^2 \left(50\right)\ }}{2\cdot 9.8}\approx 3.0\ m}

The position along the $x$-axis for which he reaches his maximum height is then given by

\boxed{x_{max}=x\left(t_1\right)=v_0{\mathrm{cos} \left(\theta \right)\ }t_1=\frac{v^2_0{\mathrm{sin} \left(\theta \right)\ }{\mathrm{cos} \left(\theta \right)\ }}{g}=\frac{{10}^2\cdot {\mathrm{sin} \left(50\right)\ }{\mathrm{cos} \left(50\right)\ }}{9.8}\approx 5.02\ m}

3. Find the total time of flight.

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The total time of flight $t_2$ is such that $y\left(t_2\right)=0$ and we solve for $t_2$ as follows

\begin{aligned} y\left(t_2\right)=0\ \ \ \ &\Rightarrow \ \ \ \ \ \ v_0{\mathrm{sin} \left(\theta \right)\ }t_2-\frac{1}{2}gt^2_2=0 \\ \\ &\Rightarrow \ \ \ \ \ \ \left(v_0{\mathrm{sin} \left(\theta \right)\ }-\frac{1}{2}gt_2\right)t_2=0 \\ \\ &\Rightarrow \ \ \ \ \ \ t_2=0\ \ \ \ \ or\ \ \ \ \ t_2=\frac{2v_0{\mathrm{sin} \left(\theta \right)\ }}{g} \end{aligned}

Keeping only the nonzero root, we conclude that the total time of flight is equal to

\boxed{t_2=\frac{2v_0{\mathrm{sin} \left(\theta \right)\ }}{g}=\frac{2\cdot 10\cdot {\mathrm{s}\mathrm{in} \left(50\right)\ }}{9.8}\approx 1.56\ s}

4. How much time does he spend at a height greater or equal to $\displaystyle{\frac{3}{4}y_{max}}$?

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The time spend above $75\%$ of $y_{max}$ is given by solving $y\left(t\right)=0.75\cdot y_{max}=2.25\ m$ and subtracting the two roots from each other.

\begin{aligned} y\left(t\right)=0.75\cdot y_{max}\ \ \ \ &\Rightarrow \ \ \ \ \ v_0{\mathrm{sin} \left(\theta \right)\ }t-\frac{1}{2}gt^2=0.75\cdot y_{max} \\ \\ &\Rightarrow \ \ \ \ \ 10\cdot {\mathrm{sin} \left(50\right)\ }t-\frac{1}{2}\left(9.8\right)t^2=0.75\cdot 3 \\ \\ &\Rightarrow \ \ \ \ \ -4.9t^2+7.66t-2.25=0 \\ \\ &\Rightarrow \ \ \ \ \ \left\{ \begin{array}{c} t_3\approx 0.40\ s \\ t_4\approx 1.17\ s \end{array} \right. \end{aligned}

The time spend above $3y_{max}/4$ is given by the difference $\mathrm{\Delta }t=t_4-t_3$ and is equal to

\boxed{\mathrm{\Delta }t=1.17-0.40=0.77\ s}

Therefore, Tony Parker spends $50\%$ of the time of flight above $0.75\cdot y_{max}$. That is, he spends $50\%$ of the time at a height greater than $0.75\cdot y_{max}$ of the motion which gives the impression that he is floating in midair.

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