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PROFESSOR: So in
this example, we

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want to hit an apple hanging
from a tree with a projectile.

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And the main point
of the problem

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is to figure out at
what angle to the ground

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should we aim our projectile
when we fire it off

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in order to hit the apple.

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And in this example,
we're assuming

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that the apple drops from
the tree at the same instant

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that we fire the projectile.

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So in this drawing, let's just
put some dimensions in here,

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we'll assume that
the apple starts out

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at a height, h,
above the ground.

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That the horizontal
distance from where

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the projectile starts to
the apple is a distance, d.

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That the projectile starts at a
distance, s, above the ground.

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And that we fire the projectile
off with an initial velocity,

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v0, at an angle theta, sorry,
theta not, with respect

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to the horizontal.

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And let's define our origin to
be right here on the ground,

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directly below where
the projectile begins.

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We need to define our
coordinate system,

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so the i-hat direction will
be horizontally to the right,

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and the j-hat direction
will be vertically upwards

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with an origin at this point.

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So the question that
we have is, what

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should our angle,
theta0 b, in order

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to hit the apple,
which starts falling

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from the tree at
the same instant

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that we fire our
projectile, and we

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want to hit the apple
before it hits the ground.

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OK so to work this out,
we need to consider

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the kinematics of two separate
objects, our projectile

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and the apple.

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So let's begin with the apple.

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So for our apple, the apple
drops vertically straight

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downwards, so its horizontal
position throughout its motion

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is just unchanged.

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So the x-coordinate of the
apple as a function of time

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is just d, it's just a constant.

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It's y-coordinate, it
starts out at a height, h,

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and then it drops due to
the acceleration of gravity,

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and so that's given by
minus 1/2 gt-squared.

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It's a minus because it's
falling in the minus j-hat

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direction.

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Now for our projectile,
let's call it a bullet.

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For our bullet, the
x-coordinate of the bullet,

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it starts out at the
origin, at x equals zero.

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And its initial motion, it
has an initial velocity, v0,

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in this direction.

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The horizontal
component of that is

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v0 times the cosine of theta0.

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And so it's displacement
of time t is v0 t times

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the cosine of theta 0.

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And there's no
horizontal acceleration,

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so that completes
our x-coordinate

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as a function of time.

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For our y-coordinate,
we start out

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at a height, s,
above the ground.

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There is the change
in the y-coordinate

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due to the initial velocity,
which has component v0 sine

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theta0, so that's plus
v0 t sine theta 0.

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And there's also a
vertical acceleration

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due to gravity, which gives
us minus 1/2 gt-squared.

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OK so that's the kinematics
of the apple and the bullet.

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Now for there to be a hit
at time t equals capital-t,

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the coordinates of the
apple and the bullet

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have to be the same at
that collision time.

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So for a hit at t
equals big T, we

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require that the x-coordinate
of the bullet at time big T

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is the same as the
x-coordinate of the apple.

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And likewise, that the
y-coordinate of the bullet

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is equal to the
y-coordinate of the apple.

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OK so let's look at
each of these in turn.

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For our x-coordinate, the
x-coordinate of the bullet

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is v0 capital-T times
the cosine of theta 0.

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And that has to be equal
to the x coordinate

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of the apple, which is just d.

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So notice that I can solve this
for the time of the collision,

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capital-t, and that's
just d minus v0 times

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the cosine of theta 0.

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Now for our y-coordinate, the
y-coordinate of the bullet

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is s plus v0 capital-T
time is the sine of theta

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not minus 1/2 gt
squared and that's

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equal to the y-coordinate
of the apple, which

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is h minus 1/2 gt squared.

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All right, now I
can rearrange that.

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Notice that I have a minus
1/2 gt squared on both sides,

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so those cancel out.

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I can rearrange that to write
v0 times capital-T times

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the sine of theta 0
is equal to h minus s,

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so that the sine of theta 0
is equal to h minus s over v0

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capital T. But remember we
solved for capital-T here,

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so I can substitute that in.

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So that gives me h minus
s over v0 times 1 over t,

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which is v0 cosine
theta 0 over d.

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And so I can rewrite that
as h minus s over d times

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to cosine of theta 0.

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So rewriting that, I have
on the left hand side,

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sine of theta 0 over
cosine of theta 0

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is equal to h minus s over d.

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And notice on the left
hand side, sine over cosine

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is just tangent, so this is
just the tangent of theta 0.

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Now think about what that means.

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If I draw a right triangle
where this is theta 0,

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then this is h minus
s and this is d.

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And so this is the same
geometry we have here,

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if we drew the
triangle like this.

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So theta 0 is just the angle
to the location of the apple

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just before it drops.

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So what this
calculation tells us

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is that the correct thing to
do if we want to hit the apple,

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if we know the apple is going
to drop at the same instant

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that we fire, then we
should aim at the location

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that the apple is
at that instant.

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We shouldn't try and lead
the apple and fire below it

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or fire above it, we should
aimed directly at the apple.

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And what the
kinematics shows us is

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that both the
bullet and the apple

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fall vertically down
at the same rate.

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And so that will
give us a collision.

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Now it's worth thinking
about two different cases,

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which I'm not going
to solve for you here.

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The first is, what if
the apple didn't drop?

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Suppose the apple just stayed
in the tree and I fired.

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If I aimed directly at the
apple and the apple didn't drop,

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then I would miss the apple,
because the bullet would

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drop as it went across
this distance, d.

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Now of course if the
apple were big enough,

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or if the bullet were
flying fast enough,

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if v 0 was fast
enough, then it might

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be that the amount
of the bullet dropped

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wouldn't be bigger than
the thickness of the apple,

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so I might still
graze the apple,

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but I wouldn't hit
the apple dead center.

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So if I knew that the
apple wasn't going to drop

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and I wanted to
hit it dead center,

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I would have to choose some
different angle, theta 0.

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We can sort of see intuitively
that that angle would

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have to be a little bit steeper
so that the bullet would drop

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and then hit the apple.

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And so the way we
would solve that is

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in our original
kinematic equations.

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For the motion of the
apple, x would still

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be a constant, d, but
if the apple wasn't

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dropping then y of the apple
would also be a constant.

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We wouldn't have the second
term, we would just have y

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equals h.

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And then I would have to
solve for a collision.

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The other case to
think about, that I'd

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like you to think about,
is what if the apple

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begins dropping for a short
interval before we fired.

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So that at the time that
we fire our projectile,

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the apple is already dropping.

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Well that's equivalent
to saying that at time

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equals 0, when I
fire, that the apple

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has some initial
velocity, whatever

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velocity it's picked
up by dropping

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and whatever interval
it was dropping before.

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So now, my kinematics
for my apple

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in the vertical
direction, I would

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have an additional term
here, a minus v0 t,

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where v0 is that
initial velocity.

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And that's a
different v0, I should

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have used a different symbol.

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It's a different v0 than
the v0 of the bullet,

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but there would be an initial
velocity, falling velocity,

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of the apple that
I'd have to consider.

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So the particular details of
what's happening with the apple

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change the kinematics,
and it's worth

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thinking about how that
would change your answer.