Mr. P: Good morning!
Today we are going to use the equations of projectile motion
to derive the range equation.
Bo: Hey guys.
Billy: Hey Bo!
Bobby: Hi Bo.
- ♫ Flipping Physics ♫
Mr. P: Remember, the range of a projectile is defined
as the displacement in the x direction,
when displacement in the y direction equals a zero.
And, the range equation is the range equals the magnitude
of the initial velocity squared, times the sine of
two times the launch angle, all divided by
the acceleration due to gravity.
Alright, let's begin deriving the range equation.
Billy, could you please break the
initial velocity into its components?
Billy: I thought that when we used the range equation
we used the initial angle and the
magnitude of the initial velocity.
We did not resolve the initial velocity into its components,
isn't that right?
Mr. P: Correct, you do not need to break the initial velocity
into its components in order to use the range equation.
However, in order to derive the range equation, you do.
Billy: Oh right, my bad, dawg.
Sine theta equals opposite over hypotenuse,
therefore sine theta initial equals velocity initial
in the y direction over velocity initial,
and multiplied by the velocity initial, gives us that
velocity initial in the y direction equals
the velocity initial times the sine of the initial angle.
And cosine theta equals adjacent over hypotenuse,
therefore cosine theta initial equals velocity initial
in the x direction over velocity initial,
and again, multiplying by velocity initial gives us
velocity initial in the x direction equals
velocity initial times the cosine of theta initial.
But we don't have any numbers to
plug into the equations, do we?
Mr. P: That is correct, we do not have any numbers,
because we're not solving a number problem here,
we're actually using all variables because we're
deriving an equation, we're deriving the range equation.
Bo, could you please list everything we know
in the x and y directions?
Bo: In the x-direction, we know the projectile will
move at a constant velocity, so the initial velocity
which is found in the x-direction, which is
the velocity initial times the cosine of the launch angle,
is equal to the velocity in the x-direction.
We're solving for the displacement in the x-direction
or the range, so R equals question mark.
In the y-direction we know the displacement
will be equal to zero, and the acceleration is
negative 9.81 meters per second squared.
Mr. P: Actually Bo, we should not use numbers when
deriving equations, we should use only variables.
Bo: Okay, then the acceleration in the y-direction
equals negative little g, the acceleration due to gravity.
Mr. P: Bobby, please start solving the problem in the x-direction.
Bobby: What am I solving for?
I didn't think we were doing a problem.
Mr. P: Ah, the problem is derive the range equations.
So what do you think then, we should be solving for,
in the x-direction?
Bobby: So we should solve for the displacement
in the x-direction, or the range.
We know the velocity in the x-direction equals the
displacement in the x-direction, divided by
the change of time, multiply both sides by the
change in time to get the x displacement, or the range,
equals the change of time times
the velocity in the x-direction.
Bo: And we have an equation for the
velocity in the x-direction.
It equals the initial velocity times
the cosine of the launch angle.
Bobby: Yeah, so the range equals the change in time,
times the velocity initial,
times the cosine of the initial angle.
Mr. P: Right, so now we have an equation for range,
in terms of the change in time, the initial velocity,
and the initial launch angle.
Let's pause now, switch to the y-direction,
solve for change in time,
substitute what we get back into this equation for range.
Billy, could you please do that?
Billy: We can use the uniformly accelerated motion equation,
displacement in the y-direction equals
the initial velocity in the y-direction,
times the change in time, plus 1/2 times
the acceleration in the y-direction,
times the change in times squared.
Change in Times Square, get it?
Bobby: I get it, but it's not funny.
Bo: It's kinda funny.
Billy: Thanks, Bo.
Mr. P: Okay, that's a bit weird.
Can we get back to the problem, please?
Billy: The displacement in the y-direction is zero,
so the whole equation equals zero.
So we can divide the equation by the change in time
to get zero equals velocity initial in the y-direction
plus 1/2 times the acceleration in the y-direction
times the change in time.
Subtract 1/2 times the acceleration in the y-direction
times the change in time from both sides,
and we get the initial velocity in the y-direction
equals to the negative of 1/2 times the acceleration
in the y-direciton times the change in time.
We can substitute in the negative g
for the acceleration in the y-direction,
and the negative cancels out, which means we are
left with 1/2 times the acceleration due to gravity
times the change in time.
Now multiply both sides by two, to get
two times the velocity initial in the y-direction
equals the acceleration due to gravity,
times the change in time, and then
divide both sides by the acceleration due to gravity
to get the change in time equals
two times the initial velocity in the y-direction,
divided by the acceleration due to gravity.
Oh, and you could substitute in velocity initial
times the sin of the initial angle
for the velocity initial in the y-direction,
and we get the change in time equals
two times the velocity initial
times the sine of the initial angle,
all divided by the acceleration due to gravity.
Mr. P: And now you can substitute that equation for change
in time back into our equation for range.
And we get that the range is equal to
two times the magnitude of the initial velocity
times the sine of the initial angle,
divided by the acceleration due to gravity,
all multiplied by the magnitude of the initial velocity
times the cosine of the launch angle.
And we can re-arrange it a little bit so it looks like this.
The range of a projectile is equal to
the magnitude of the initial velocity squared
times two times the sine of the launch angle
times the cosine of the launch angle,
all multiplied by the acceleration due to gravity.
And, who sees it?
Billy: Uh, sees what?
Bobby: I don't know.
Bo: Yeah, I'm not seeing what we're supposed to be seeing.
Mr. P: Yeah, that's okay, in this class
generally we don't require that you remember,
or even have learned the double angle formula
or other trick functions like it.
Bo: Double Angle Formula?
Bobby: Oh yeah, two times the sine of theta
times the cosine of theta, equals
the sine of two times theta.
Billy: I don't think I ever learned that.
Mr. P: Again, you do not need to know the Double Angle Formula,
two times the sine of theta, times the
cosine of theta is equal to the sine
of two theta, in order to use the range equation.
We do need, however, in order to do this very last step
in deriving the range equation.
We're going to substitute in the
sine of two times theta initial.
The range of a projectile equals
the magnitude of the initial velocity squared
times the sine of two times the launch angle
divided by the acceleration due to gravity.
Bo: g
Bobby: is
Billy: positive!
Mr. P: Of course, we all remember that the acceleration
due to gravity is a positive number.
That, is the derivation of the range equation.
Please remember the range is defined as the
displacement in the x-direction of a projectile,
and the overall displacement in the y-direction
is equal to zero, V sub i is the magnitude
of the initial velocity, and theta i is the initial angle
or the launch angle of the projectile,
and g, of course, is positive.
Thank you very much for learning with me today,
I enjoyed learning with you.