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Angolo ottimale per un proiettile parte 1

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Let's say we're going to shoot some object into
the air at an angle.
Let's say its speed is s and the angle at which we shoot
it, the angle above the horizontal is theta.
What I want to do in this video is figure out how far
this object is going to travel as a function of the angle and
as a function of the speed, but we're going to assume that
we're given the speed.
That that's a bit of a constant.
So if this is the ground right here, we want to figure out
how far this thing is going to travel.
So you can imagine, it's going to travel in this parabolic
path and land at some point out there.
And so if this is at distance 0, we could call this distance
out here distance d.
Now whenever you do any problem like this where you're
shooting something off at an angle, the best first step is
to break down that vector.
Remember, a vector is something that has magnitude
and direction.
The magnitude is s.
Maybe feet per second or miles per hour.
And the direction is theta.
So if you have s and theta, you're giving me a vector.
And so what you want to do is you want to break this vector
down into its vertical and horizontal components first
and then deal with them separately.
One, to help you figure out how long you're in the air.
And then, the other to figure out how far
you actually travel.
So let me make a big version of the vector right there.
Once again, the magnitude of the vector is s.
So you could imagine that the length of this arrow is s.
And this angle right here is theta.
And to break it down into its horizontal and vertical
components, we just set up a right triangle and just use
our basic trig ratios.
So let me do that.
So this is the ground right there.
I can drop a vertical from the tip of that arrow to set up a
right triangle.
And the length of the-- or the magnitude of the vertical
component of our velocity is going to be this
length right here.
That is going to be-- you could imagine, the length of
that is going to be our vertical speed.
So this is our vertical speed.
Maybe I'll just call that the speed sub vertical.
And then, this right here, the length of this part of the
triangle-- let me do that in a different color.
The length of this part of the triangle is going to be our
horizontal speed, or the component of this velocity in
the horizontal direction.
And I use this word velocity when I specify
a speed and a direction.
Speed is just the magnitude of the velocity.
So the magnitude of this side is going to be speed
horizontal.
And to figure it out, you literally use
our basic trig ratios.
So we have a right triangle.
This is the hypotenuse.
And we could write down soh cah toa up here.
Let me write it down in yellow.
soh cah toa.
And this tells us that sine is opposite over hypotenuse,
cosine is adjacent over hypotenuse and tangent is
opposite over adjacent.
So let's see what we can do.
We're assuming we know theta, we know s.
We want to figure out what the vertical and the horizontal
components are.
So what's the vertical component going to be?
Well the vertical component is opposite this theta.
But we know the hypotenuse is s, so we could use sine
because that deals with the opposite and the hypotenuse.
And the sine function tells us that sine of theta-- actually,
let me do this in green since we're doing all the vertical
stuff in green.
Sine of theta is going to be equal to opposite, which is
the magnitude of our vertical velocity.
So the opposite side is this side right here, over our
hypotenuse.
And our hypotenuse is the speed s.
And so if we want to solve for our vertical velocity or the
vertical component of our velocity, we multiply both
sides of this equation by s.
So you get s sine of theta is equal to the vertical
component of our velocity, s sine of theta.
And now for the horizontal component we do the same
thing, but we don't use sine anymore.
This is now adjacent to the angle.
So cosine deals with the adjacent side and the
hypotenuse.
So we could say that the cosine of theta is equal to
the adjacent side to the angle, that is the horizontal
speed, over the hypotenuse.
The hypotenuse is this length right here, over s.
So if we want to solve for the horizontal speed or the
horizontal component or the magnitude of the horizontal
component, we'd just multiply both sides times s.
And you get s cosine of theta is equal to
the horizontal component.
So we now know how fast we are travelling in this direction,
in the horizontal component.
We know that that is going to be s cosine of theta.
And we know in the vertical direction-- let me do that in
the vertical direction, the magnitude is s sine of theta.
It is s sine of theta.
So now that we've broken up into the two components, we're
ready to figure out how long we're going to be in the air.