In this section, we look at the path
or trajectory of a projectile and we also determine its range.
There's a bit of algebra coming up,
so take it slowly and don't be reluctant to pause
or to play bits again.
It's always a good idea to do the algebra yourself on a piece of paper.
Remember, doing is much better than watching.
Let's start with our general equations for constant acceleration.
As usual, we neglect air resistance and set a x equals zero.
Let's also make the projectile start from the origin -
so we have x zero equals zero and y zero equals zero.
For projectiles, we set a y equals minus g.
So our equations for x and y motion are x equals v x zero t
and y equals v y zero t minus a half g t squared.
That's two equations, but we want the trajectory, y of x.
So we can rearrange equation one to eliminate t.
Substitute for t in two gives this equation for y as a function of x.
We can write it like this to make it clear that y is a quadratic function of x.
y of x is a parabola.
Let's be clear - what is y of x?
Y is the height above the launch point
and x is the horizontal displacement from the launch.
So the curve, y of x, is the path in space.
It is the trajectory followed by the projectile and that trajectory is a parabola.
Let's look closely at the terms in this quiz.
Let's recap from the quiz.
The coefficient of x gives the initial slope,
which is the direction of the initial velocity vector, v zero.
If we launch at an angle theta above the horizontal,
10 theta equals v y zero over v x zero.
The coefficient of x squared is minus g over 2 v x zero squared.
The minus sign tells us that the parabola curves downwards - well, things fall down.
Also, it tells us that if v x is large, it doesn't curve down so much.
Let's now use the trajectory to interpret our equations for x of t and y of t.
Equation Roman i says that x increases at a constant rate;
equal increasing x in equal times.
So the velocity component in the x direction doesn't change; remember,
we're neglecting air resistance.
In Equation ii, the first term says that the projectile starts going upwards - well,
if v y zero is positive.
The second term tells us that the upwards component of v decreases;
that is, it's accelerating downwards.
Of course, if v y zero were less than zero -
if we throw it downwards - then it just keeps going faster
and faster with time in the downwards direction.
This curve in space - y as a parabolic function of x - is the trajectory
for projectiles near the earth's surface if air resistance is negligible.
When gravity is the only non-negligible force acting, we say an object is in free fall.
To get astronauts used to free fall, NASA has a plane -
it's nicknamed the "vomit comet" -
that flies along such a parabola for about 25 seconds.
The humans inside travel along the same parabola
with gravity the only force acting on them.
They seem to be floating.
In fact, they are accelerating downwards at 9.8 meters per second per second,
but so is the plane; so that they don't accelerate with respect to the plane.
More about free fall when we look at gravity.
Here is another important question.
How far does a projectile travel?
How does its range depend on the magnitude and the direction of the initial velocity?
Let's do an experiment first.
These projectiles are drops of water.
They're small, so the approximation
that the air resistance is negligible is a rather crude one.
Let's see how the range depends on angle.
If I make the angle very small, then they're not in the air for very long.
So even though they've got a large x velocity, they don't travel very far.
If I make the angle large - close to vertical -
then they're in the air for a long time,
but they don't travel very far 'cause they've got the small x component of velocity.
So maximum range will be some angle between 90 degrees and zero.
Our object; find out what it is.
Here's the equation for the trajectory.
To find the range R, we want to know when it hits the ground - when y equals zero.
So let's substitute zero for y and R for x.
We cancel R and we move the second term to the left-hand side, then cancel v x zero.
Then we rearrange to have an expression for the range R.
Well, it's no surprise that the range is proportional to v x zero;
it's also proportional to v y zero,
because the faster you go up, the longer you stay up.
Next, we substitute for the components of v zero.
And we can simplify that because of a trigonometric identity
- sin two theta equals two sin theta times cos theta.
So here's our expression for the range.
Note that it goes as the square of the takeoff speed.
Twice as fast takes you four times as far.
Why squared?
Well first, more v zero gives it more x velocity.
But it also gives it more y velocity, which keeps it above the ground longer.
So what angle gives maximum range?
The maximum value of sin two theta is one, which occurs when theta equals 45 degrees.
Well, that's roughly in agreement with our experiment - only roughly,
because here the air resistance is small, but not zero.
Well, that's enough material to do a range of problems on projectiles.
So let's do the next quiz before we return to look at circular motion.
Prof. Joe WolfeProfessor
Dr Elizabeth J. AngstmannFirst Year Physics Director
Mr Sebastian FrickeTeacher and laboratory manager
