if placed by 12PM EST

Free US Shipping on orders over $25

Are you a football fan? Are you ever amazed at how high and far the kickers and punters are able to launch the ball? Did you know you can use physics to figure out how high the ball goes? From home? Yep!

To cut to the chase, here’s how you can determine height from hang-time. Pull out your stopwatch and record the hang-time. In other words, you press start when the kicker’s foot connects with the ball and then you press stop when the ball hits the ground.

Multiply the hang-time by 2 and square the result, and that is the approximate height in units of feet! Pretty cool, huh!

Here’s how to figure this out using Newton’s laws! Isaac Newton’s second law proposed that the acceleration experienced by an object is proportional to the force acting upon it. We write it this way,

*F = ma*,

where *F* is the force, *a* is the acceleration, and *m* is the mass. For a football traveling through the air, there are two forces acting on the football: gravity and drag. For simplicity, we’ll just look at gravity and ignore drag.

The force of gravity on the football is equal to its mass times the coefficient of gravity, *g* = 9.8 *m*/*s^2*. Thus, Newton’s second law gives us −*mg* = *ma*, or *a* = −*g*.

Now we add some hints of calculus. First, we point out that average acceleration is defined as the change in velocity per unit time. Since the acceleration is constant, the average acceleration is just the acceleration. Thus, we can write

*a* = [*v(t)* − *v(0)*] / *t*,

where *v(t)* is the velocity at a certain time *t*, and *v(0)* is the initial velocity of the football leaving the kickers foot. Solving to for *v(t)*, we arrive at *v(t)* = *v(0)* −*gt*.

Up until now, we haven’t really talked about the direction the ball is kicked. But as we all know, the angle of the kick will definitely affect the hang-time.

If the ball is kicked straight-up, then the hang-time will be huge (but the ball won’t go very far). If the ball is kicked straight-forward the hang-time will be essentially zero. See the following graphic which shows the trajectory of a football kicked with the same initial velocity at different angles.

The football achieves different heights because the force of gravity only acts in the vertical direction. Thus, the force of gravity only affects the vertical component of the motion.

Therefore, in the equation above, we’re really only talking about the vertical component of the velocity, *v(t)* = *v(0)* − *gt*. The horizontal component of the velocity doesn’t change (as long as we’re ignoring air resistance). We can write this as *vx(t)* = *vx(0)*.

This point is key because we want to determine something about the height the football travels, not the distance. The final step is to determine the vertical position of the football as a function of time, *y(t)*. This can be done by using full blown calculus and noting that the velocity is the derivative of the position,

*vYY(t)* = *dy(t)/dt *.

If this seems complicated – well, it is calculus after all. But hang with me for just another second! What we need to do is determine a function *y(t)* that has a particular *v*YY*(t)* that satisfies the equation above. It turns out that the solution is not very complicated,

*y(t)* = *y(0)* + *vYY(0)t* − *gt^2/2*,

where *y(0)* is the height at which the kickers foot hits the ball – essentially zero for a kick-off with a tee. So, we’re left with the following equation for the vertical position of the ball

*y(t)* = *vYY(0)t* − *gt^2/2*,

Now we have to get a bit tricky. After the kick the ball will go up and eventually it will come back down and hit the ground. At that time (let’s call it *t* = *T*) the vertical position will be zero, *y(T)* = 0. This is critical piece of information because we can use it to find *vYY(0)*,

*yYY(0)* = *gT/2*

So, we now have

*y(t)* = [*gt(T* − *t)*]/2.

The final step is to figure out the time (let’s call it *t* = *t*MAXMAX) at which *y(t)* is maximum. But we can be tricky again. When the ball is at its highest point, the velocity in the vertical direction will be zero. We can write this *v(t*MAXMAX*)* = 0 and use it to work out that

*t*MAXMAX = *T*/2.

Finally, we have

ℎ = *y*(*t*MAXMAX) =* gT*^2 / 8 .

where ℎ is the maximum height achieved by the football. We can simplify this so it’s easy to calculate during a game in units of feet. (I won’t do it here, but you convert *g* = 9.8 m/s^2, to ft/s^2, and then calculate ROOT*g*/8 noting that* gT*^2/8 = (ROOT(*g*/8)*T*)22.

*h* ~ (2T)^2 ft/s^2

Now, during the next football game, you can pull out your stopwatch, record the hang-time, and easily estimate the kick-off height. In other words, you press start when the kicker’s foot connects with the ball and then you press stop when the ball hits the ground. Multiply the hang-time by 2 and square the result and that is the approximate height units of feet! Amazing!

You can also use this trick for determining the height of a field goal kick. The result will be a little less accurate because the downward descent of the football will slow a bit when it hits the net behind the goal. But the upside is that the field goal standards are 30 ft tall, so you can see how your estimate of the height compares to reality.