## Hang Time of a Leap

Have you ever watched an athlete leap into the air and they appear to float for a long time.? We can watch Ballet Dancers and Basketball Players do this; but what is going on?

This is a commented on phenomena that shows the subtlety in analyzing physics equations. It has been said that a good athlete appears to float in mid-air. The equations of motion for free fall hint at this but let us play with the equations to see this more clearly.

We begin by describing the motion in just the vertical direction. If we have a good athlete that is jumping straight up, which is common in basketball and ballet, and happens to be two of the sports where these phenomena are commonly described.

We are taking a person jumping as our motion. So the force they apply will be converted to an initial velocity. This velocity is the input that defines the rest of the motion. The only other force acting is gravity. So we have a constant acceleration we can calculate the height using the following equation.

$$h=\frac{1}{2}a t^2+v_0 t+x_0$$

We will adapt this for our purposes by defining some reference points

$$x_0=0$$

$$a=-g=-9.8 m/s$$

This gives us this format we can use to manipulate these equations:

$$h=-\frac{1}{2}g t^2 +v_0 t$$

If we graph height as a function of time we get this shape to the equation.

Figure 1: This shows the height of an object given an initial velocity of 5 m/s and g of 9.8 m/s^2.

We want to recognize this quadratic equation is parabolic in time and we can use the vertex solution to define the time it takes to reach the peak height.

$$t_{peak}=\frac{v_0}{g}$$

Because this is parabolic it is symmetric so our total time of flight is just twice the time it takes to get to the vertex.

$$t_{flight}=\frac{2 v_0}{g}$$

Using the time for peak height we solve for the peak height

$$h_{peak}=-\frac{1}{2}g \left(\frac{v_0}{g}\right)^2+v_0 \left(\frac{v_0}{g}\right)$$

$$h_{peak}=\frac{v_0^2}{2g}$$

So we have an idea of how high our jumper will jump given an initial velocity, and how long they will remain in the air. Let us now calculate the length of time they will be higher than half the maximum height. We can do this by solving for one half of the peak height. This will give us two solutions which are the time the jumper first reaches half of peak height and the time the jumper falls back down to half the height. The difference between these two times is the total length of time the jumper will be above half the peak height.

$$\frac{1}{2}h_{peak}=-\frac{1}{2}g t^2 + v_0 t$$

$$0=-\frac{1}{2}g t^2 + v_0 t – \frac{1}{2}h_{peak}$$

$$t_{hang}=\frac{-v_0\pm\sqrt{v_0^2-4\left(-\frac{1}{2}g\right)\left(-\frac{1}{2}h_{peak}\right)}}{-g}$$

We can substitute what we found for the peak height back in and we get

$$t_{hang}=\frac{-v_0\pm\sqrt{v_0^2-\frac{v_0^2}{2}}}{-g}$$

$$t_{hang}=\frac{-v_0\pm\sqrt{\frac{1}{2}v_0^2}}{-g}$$

The time the jump is above the half peak height is the difference between the two solutions.

$${\Delta} t=\frac{-v_0-\frac{\sqrt{2}}{2}v_0+v_0 – \frac{\sqrt{2}}{2}v_0}{-g}=\frac{\sqrt{2}v_0}{g}$$

The ratio of this time to the total time of flight is

$$\frac{{\Delta}t}{t_{flight}}=\frac{\frac{\sqrt{2}v_0}{g}}{\frac{2 v_0}{g}}=\frac{\sqrt{2}}{2}\approx 0.7071$$

So about 71% of the flight time is in the upper half of the height.

And in the animation below you can see this in the motion of the jumping ball.

Figure 2: A full leap at 5 m/s initial velocity and a g of 9 m/s^2 animated against the graph of height as a function of time.

So athletes are not floating they are just moving in a constant acceleration. Their vertical motion follows rules. The illusion of floating comes from our perception of the time they are in the air.

### Code Used To Generate the Images

The graphs and animation were produced in Mathematica 9. The animation for the gif was produced in Mathematica 9.

##### Mathematica Code for Figure 1:

v0 = 5;
h[t_] := -(1/2) g t^2 + v0 t;
g = 9.8;
Plot[h[t],
{t, 0, (2 v0)/g},
AxesLabel ->{"Height - m", "Time - s"},
PlotRange ->{0, v0^2/(2 g)},
PlotStyle -> {Blue}]


##### Mathematica Code for the Animation in Figure 2:

v0 = 5;
g = 9.8;
h[t_] := -(1/2) g t^2 + v0 t;

Animate[
Show[
Plot[
h[t],
{t, 0, (2 v0)/g},
AxesLabel -> {"Height - m", "Time - s"},
PlotRange -> {0, v0^2/(2 g) + 0.4},
PlotStyle -> {Blue, Thick}],
Plot[
v0^2/(4 g),
{t, 0, (2 v0)/g},
PlotStyle -> {Green, Dashed, Thick}],
Graphics[
{
PointSize[0.05],
Point[
{v0/g, h[t]}
]
}
]
], {t, 0, (2 v0)/g}
]

##### Mathematica Code for Creating Individual Frames of Figure 2:


v0 = 5;
g = 9.8;
h[t_] := -(1/2) g t^2 + v0 t;

v = Table[
Show[
Plot[
h[t],
{t, 0, (2 v0)/g},
AxesLabel -> {"Time - s", "Height - m"},
PlotRange -> {0, v0^2/(2 g) + 0.4},
PlotStyle -> {Blue, Thick}
],
Plot[
v0^2/(4 g),
{t, 0, (2 v0)/g},
PlotStyle -> {Green, Dashed, Thick}
],
Graphics[
{
PointSize[0.05],
Point[
{v0/g, h[t]}
]
}
]
],
{t, 0, (2 v0)/g, (2 v0)/(50 g)}
]

Export["hang_time_10.gif", v, DisplayDurations -> "0.05"]