91影视

The time the athlete in air is,$t=1.00s$.

The maximum height of the athlete is,$Y_{max}$.

The object's highest vertical position is its greatest height.

An object's maximum height is determined by the starting velocity, launch angle and gravity-induced acceleration.

The equation of the speed of the athlete is expressed as-

$v^2=v_0^2-2g\frac{Y_{max}}{2}$ 鈥�1)

Here,$Y_{max}$is the maximum height, v is the speed of the athlete, g is the acceleration due to gravity and$v_0$is the initial speed.

The equation of the maximum height can also be written with the help of the equation of the kinetic and potential energy

$K_0+U_0=K+U$

Here,$K_0$are$U_0$the initial kinetic and potential energy$K$and $U$are the final kinetic and potential energy.

The above equation can be written as-

$\frac{m{v}_0^2}{2}+0=0+mg{y}_{max}$

Here, m is the mass of the athlete

The above equation can be reduced as,

$$\begin{gathered} v_0^2=2g{y}_{max} \\ {y}_{max}=\frac{v_0^2}{2g} \end{gathered}$$ 鈥�2)

Substituting the equation 2) in equation 1),

$$\begin{gathered} v^2=v_0^2-g\left(\frac{v_0^2}{2g}\right) \\ =v_0^2-\frac{v_0^2}{2} \\ =\frac{v_0}{\sqrt{2}} \end{gathered}$$

The equation for obtaining the$t_{half}$is expressed as,

$$\begin{gathered} g{t}_{half}=\frac{v_0}{\sqrt{2}}...3) \\ {t}_{half}=\frac{v_0}{\sqrt{2}g} \end{gathered}$$

As the athlete has spent double of the $t_{half}$above the required height$\frac{Y_{max}}{2}$, then the equation of the total height is,

$$\begin{gathered} t_{above}=t_{half}+t_{half} \\ =2t_{half} \end{gathered}$$

Substituting equation 3) in the above equation,

$$\begin{gathered} t_{above}=\frac{2v_0}{\sqrt{2}g} \\ =\frac{\sqrt{2}{v}_0}{g} \end{gathered}$$$$\begin{gathered} t_{above}=t_{half}+t_{half} \\ =2t_{half} \end{gathered}$$

The equation of the total time can be expressed as,

localid="1655801054170" $$\begin{gathered} gt=v_0 \\ t=\frac{v_0}{g} \end{gathered}$$

The equation of the time from the floor is expressed as,

localid="1655801059257" $$\begin{gathered} t_{fromfloor}=\frac{v_0}{g}-\frac{v_0}{\sqrt{2}g} \\ =\frac{v_0\left(\sqrt{2}-1\right)}{\sqrt{2}g} \end{gathered}$$

localid="1655801065996" $$\begin{gathered} \mathrm{Hence},\mathrm{the}\mathrm{ratio}\mathrm{of}\mathrm{the}{t}_{fromfloor}/t_{fromfloor}\mathrm{is}\mathrm{expressed}\mathrm{as}- \\ \frac{t_{above}}{t_{fromfloor}}=\frac{{\displaystyle \frac{2v_0}{\sqrt{2}g}}}{{\displaystyle \frac{v_0\left(\sqrt{2}-1\right)}{\sqrt{2}g}}} \\ =\frac{2}{\sqrt{2}-1} \\ =4.8 \end{gathered}$$

Thus, as the time above is times the time from the floor, then the athlete will seem to hang in the air.