How many times has this happened to you: you're on the roof of a tall building
and you happen to have some small projectile in your hand,
and you think to yourself, "I'll bet I could throw it all the way to that river across the street."
And without stopping to engage your better judgment you let it fly...
...only to come up short and see your projectile splatter on a riverside bike path.
And as you slink away hoping nobody noticed, you think maybe you didn't throw it at the
best angle to maximize the distance, and you realize that although a 45 degree angle would
be optimal on level ground, that's probably not the case when you're high off the ground,
and the answer probably depends not only on the height but on how hard you throw because if the
height is large relative to the strength of the throw then the small amount of air time
you gain by throwing it higher is not enough to compensate for the loss of horizontal velocity.
So you go home and you sit at your desk, define some variables, draw some diagrams,
work out the formula for the time spent in the air, multiply that by the constant
horizontal velocity to get the distance, and then you find the maximum of that function
by simply taking its derivative setting it to zero and solving for the launch angle.
And then you realize that that last step seems impossible and you've reached a dead end. Well,
if you're like me, it's happened at least once, and in that case this video is for you.
To figure this one out we need a change of perspective. Instead of tracking the position
of the projectile, let's take a close look at its velocity. We can represent velocities with
vectors which point in the direction of motion and whose length represents how fast it's moving.
At the start, the vector will look like this, and as it travels along with the projectile's
parabolic path, it changes continuously due to gravity until it hits the ground at a larger speed
and a downward angle. To be clear, I'm assuming ideal conditions: ignoring air resistance,
assuming the curvature of the earth is negligible, and sticking to Newtonian physics. Also, I'm
assuming that you can throw at the same speed at any angle which probably isn't quite true. Keeping
all that in mind, we can solve this problem without too much difficulty. First, it's useful
to think of the velocity at any point as the sum of a horizontal vector and a vertical vector since
the vertical component determines how high the projectile will be at any point in time, and the
horizontal component tells us the rate at which the projectile is moving away from the building.
Notice that the horizontal component doesn't change throughout the flight. Since gravity acts
only in the vertical direction, it can change only the vertical component, and since gravity is the
only thing changing the velocity in this ideal scenario, nothing will change that horizontal
speed. This means that the final distance can be computed by multiplying the total airtime by the
horizontal velocity. Now let's look closely at how the overall velocity changes during the flight.
If we bring all these vectors together, we can see more clearly how the velocity changes. Notice that
the line connecting the ends of these vectors is vertical. Why? Because this line represents
the total change in velocity, and again, the only thing changing the velocity is gravity.
And since gravity is constant, we can quantify this change as the gravity of earth (about 9.8
meters per second squared) multiplied by the total air time, whatever that may be.
Let's look just at this triangle. The upper side represents the velocity of the throw,
and the lower side represents the velocity when the projectile hits the ground.
Interestingly, its speed when it hits the ground doesn't depend on the angle of the throw!
That's because of the conservation of energy. The energy of motion, kinetic energy,
when the object reaches the ground must be equal to the kinetic energy of the throw plus
the potential energy that comes from the starting height -- well, at least in our idealized scenario
where no energy is lost to air resistance. So for the purposes of finding the best launch angle,
we can treat the magnitudes of the initial velocity and the final velocity as constants. We
know a lot about this velocity triangle but what does it have to do with maximizing the distance?
Well, you might notice that when the throw goes straight up so that the total distance is zero
the triangle flattens into a line. And it's the same when the throw goes straight down.
This might make you wonder if the distance is somehow related to the area of the triangle.
Well check this out: the area of the triangle is one half the base, which we can take to be this line,
times the height, which would be this line, which as we've already seen represents the horizontal
velocity. So the area of the triangle is one-half gT times Vh which differs from the distance by
a factor of one-half g which is a constant. In other words, if we find the launch angle
that maximizes the area of the triangle, we'll have the angle that maximizes the distance! So,
when is the triangle's area maximized? Well, if we now choose the final velocity as the base,
then since that's a constant we just need to find the angle that maximizes the height, which is this
line here. Notice that this line is always shorter than the initial velocity line...
...except when they're equal! So the area, and therefore the distance, is maximized when the angle
between the starting and ending velocities is 90 degrees. Take a moment to appreciate how cool that is.
If the projectile hits the ground at an angle that is rotated 90 degrees from its launch angle,
then it will have covered the maximum distance possible given the height and velocity it started
with. Now that we know this is a right triangle, the formula for the ideal angle falls right into
our lap. This small triangle is similar to the whole triangle since they have a common angle at
the top and they're both right triangles. So the angle at the bottom is our launch angle.
The tangent of that angle is opposite over adjacent, so the angle itself is
the inverse tangent of the initial velocity over the final velocity. Easy peasy, right?
So now you know if you can estimate the height of the building and make a good guess at how hard you
can throw, then you can get a pretty good idea of how to throw it to get the maximum distance.
And if you happen to have a nice tailwind that balances out the air resistance,
you might just be right about that.