Answer to Question #163586 in Mechanics | Relativity for Si

Question

In the vertical jump, an athlete starts from a crouch and jumps upward as high as possible. Even the best athletes spend little more than 1.00 s in the air (their “hang time”). Treat the athlete as a particle and let Y_max be his maximum height above the floor. To explain why he seems to hang in the air, calculate the ratio of the time he is above Y_max/2  to the time it takes him to go from the floor to that height. Ignore air resistance.

Answer 1

First, we need to get an expression for $v_o$. When the athlete reaches the maximum height, v = 0. So,

                                                               $v^2 = v_o^2-2gy_{max}$

                                                                   $0 = v_o^2-2gy_{max}$

                                                                $v_o=\sqrt{2gy_{max}}$

The time the athlete spends above $\large\frac{y_{max}}{2}$ :

For this case,  we are interested in the motion from $\large\frac{y_{max}}{2}$ up and back to $\large\frac{y_{max}}{2}$ again so

$y = y _o=\large\frac{y_{max}}{2}$

                                    $y -y_o = v_0t_{above}- \frac{1}{2}gt^2_{above}$

                                          $0= v_0t_{above}- \frac{1}{2}gt^2_{above}$

                                              $v_o = \frac{1}{2}gt_{above}$

                                                 $\therefore t_{above} = \large\frac{2v_o}{g}$

Substituting for $v_o$ we get:

                                                       $t_{above} = \large\frac{\sqrt{2gy_{max}}}{g}$

The time the athlete takes to reach $\large\frac{y_{max}}{2}$ jumping from the ground:

                                                    $y -y_o = v_0t_{below}- \frac{1}{2}gt^2_{below}$

                                                  $\frac{y_{max}}{2} = v_0t_{below}- \frac{1}{2}gt^2_{below}$

Rearranging the equation and multiplying by 2 to put the equation into the quadratic form:

                                            $gt^2_{below}-2v_ot_{below}+y_{max} = 0$

Now we have a quadratic equation with $a = g, \space b = -2v_o \space and \space c = y_{max}$

                                    $t_{below} = \large\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

                                   $t_{below} = \large\frac{-2v_o\pm \sqrt{4v_o^2-4gy_{max}}}{2g}$

                                        $t_{below} = \large\frac{-v_o\pm \sqrt{v_o^2-gy_{max}}}{g}$

Substituting $v_o = \sqrt{2gy_{max}}$ we get:

                                          $t_{below} = \large\frac{\sqrt{2gy_{max}}\pm \sqrt{2gy_{max}-gy_{max}}}{g}$

                                         $t_{below} = \large\frac{\sqrt{2gy_{max}}\pm \sqrt{gy_{max}}}{g}$

  We take the value with minus sign in middle of the numerator. The athlete pass by the point

$\large\frac{y_{max}}{2}$ twice, once when rising up and again when falling down the minus for the rising up event and the plus for the falling down event.

                                  $\therefore$ $t_{below} = \large\frac{\sqrt{2gy_{max}}\pm \sqrt{gy_{max}}}{g} = \frac{\sqrt{gy_{max}}}{g}(\sqrt{2}-1)$

     So, The ratio is:

       $\large\frac{t_{above}}{t_{below}}$ $= \large\frac{2\sqrt{2gy_{max}}}{g} * \frac{g}{\sqrt{gy_{max}}(\sqrt2 - 1)}$  $= \large\frac{2\sqrt2}{\sqrt2-1}$ $= 6.83$

The athlete spends more time above $\large\frac{y_{max}}{2}$ more than below it by a factor of 13.66, so they seem to be hanging in the air