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Course: Physics library > Unit 2
Lesson 2: Optimal angle for a projectile- Optimal angle for a projectile part 1: Components of initial velocity
- Optimal angle for a projectile part 2: Hangtime
- Optimal angle for a projectile part 3: Horizontal distance as a function of angle (and speed)
- Optimal angle for a projectile part 4: Finding the optimal angle and distance with a bit of calculus
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Optimal angle for a projectile part 3: Horizontal distance as a function of angle (and speed)
In this video, we'll finish writing the formula for distance as a function of the angle. Created by Sal Khan.
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I have two questions.
1) What does "rate" mean in this video?
2) The formula we got was d=2s^2*sin(theta)*cos(theta)/g, but when I learnt it we got it as d=s^2*sin2(theta)/g.
Can anybody explain why?
Thanks. :)(21 votes)
yes..rate means velocity!!
and sin2(theta)=2sin(theta)cos(theta)(20 votes)
this is way to advanced for me can someone explain it in a more simpler form(1 vote)
First you need to know the most basic formula d = r * t.
Sal was just saying that if you are given an initial velocity (s) and the angle (theta) you are throwing the object at, you can find the horizontal distance it is going to travel.
You need to first solve the rate and time using the information you are given, then you just plug them into the formula. You also need to know some trigonometry before you can understand sin and cos.
Solving for time:
First break up the initial velocity vector into its vertical and horizontal components.
You know that the gravity is pulling down the object in the vertical direction only, so it does not affect the horizontal component. The time the object is going to stay in the air is only going to be affected by the gravity. By using the law of sines, sine(theta) = opposite / hypotenuse, you will get sine(theta) = Sv / s. Multiply s on both sides, you will get s * sin(theta) = Sv.
Now by using the equation velocity = acceleration * time, you solve for time and you will get time = velocity / acceleration. Plug in the Sv and gravity (g) you get, you will get time = Sv / g. This only solves for the time it takes the object to stop moving upward. The time it takes moving downward is going to be the same as the time it takes to go up, because gravity is constant acceleration. So you will multiply the time you get by 2 to get the total time in the air. Plug in s*sin(theta) for Sv, you will get total time in air = (s*sin(theta)/g)*2 or 2s*sin(theta)/g.
Solving for rate:
Rate is just the velocity in the horizontal direction. You just need to solve for the horizontal component of the initial velocity vector. Using the trigonometric identity for cosine, cosine(theta) = adjacent / hypotenuse, so cos(theta) = Sh / s. Multiply both side by s, you will get s*cos(theta) = Sh. This is the rate.
Calculating distance it travels:
Now you get both your time and rate, just plug them in to the original d = r * t formula.
distance traveled = (s*cos(theta))* (2s*sin(theta)/g)
because everything is multiplied together, you can separate sin(theta) and cos(theta) from other numbers.
distance traveled = (s*2s/g)(cos(theta)*sin(theta)) or distance traveled = (2s^2/g)(cos(theta)*sin(theta)).
This is what Sal gets in the video.
I don't know if this simplifies the concept a little bit, but I hope it helps.(38 votes)
How does x and y component specify speed of the particle ?
I mean of the object is projected at an angle of 30degrees then how does its speed will be along x and y ?(4 votes)
An object thrown at an angle with an initial speed can be modeled as two separate velocities going in different directions, one vertical and one horizontal. Whether thrown at an angle with a velocity, or pushed horizontally and vertically simultaneously, it will cause the same effect on the object and will act the same way (it will have a curved path), and will be mathematically equivalent than the first method mentioned.(6 votes)
Isn't sin(theta)*cos(theta)=(1/2)*sin(2*theta) by reversing that sin(2*theta)=sin(theta+theta) and then we could use the sine addition identity and get sin(2*theta)=2*sin(theta)*cos(theta).
We have this in the distance formula so could it be written as (s^2/g)*sin(2*theta)?(4 votes)
Yes, it can indeed. Many textbooks phrase the formula like that instead, probably because it's a little easier to memorize.(1 vote)
- why is rate s(cos theta) and not just s?(2 votes)
s(cos theta) is the horizontal component of the s vector. The s vector by itself has a vertical component as well. Since s has a vertical component, it will not give us the correct value for horizontal distance. Hope that helps anyone with the same question.(3 votes)
- How would the ending equation end if the initial height wasn't 0?(3 votes)
In that case, you'd have a different value for time, and therefore a different value for Δx.(1 vote)
- If that is the formula for the horizontal distance with 'g' kept as a variable, would it be safe to say that the formula of gravity is:
g=(2s^2)(sinΘ)(cosΘ)/horizontal distance
yes? or no? If no, why is it wrong?(2 votes)- Generaly yes you can chance the formula to g=.... BUT remember that here g is constant!!! It is not the official formula for g! For more information about the official formula for gravity look at wikipedia: http://en.wikipedia.org/wiki/Gravitation(1 vote)
Totally different question but - would changing the mass of a spherical projectile affect the horizontal displacement when launched horizontally from a constant height above the ground, with the same initial velocity in the presence of air? For example, say a lighter (8 grams) and heavier (34 grams) projectile were put to test.(2 votes)
Of course it would. The reason being that a lighter object has less inertia compared to a heavier object. So, while the lighter object will slow down quicker, the heavier one will go farther due to its mass. In other words, the heavier object overcomes more of the air resistance (Though we don't consider air resistance in problems, we can't explain many stuff without air resistance) compared to the lighter object.
Hope this helped!(1 vote)
at0:31sal says that we have to find horizontal distance. why can't we use vertical distance?(1 vote)
Because "optimal angle" means the one that will make it go the furthest, not the highest.(3 votes)
- Horizontal distance is the range right?(1 vote)
- yes, range is the maximum horizontal distance traveled by a projectile.(1 vote)
0:31Video transcript
We now know how long the object
is going to be in the air, so we're ready to
figure out how far it's going to travel. So we can just go back to kind
of the core formula in all of really kinematics, all of kind
of projectile motion or mechanical physical problems,
and that's distance is equal to rate times time. Now, we're talking about the
horizontal distance. So our distance is going to be
equal to-- what's our rate in the horizontal direction? We care about horizontal
distance traveled, so our rate needs to be the horizontal
component of the velocity, or the magnitude of the horizontal component of the velocity. And we figured that out
in the first video. That is s cosine of theta. So let's write that
down right here. So our rate is s cosine
of theta. And how long will we
be traveling at this horizontal speed? Well, we'll be going at
that speed as long as we are in the air. So how long are we in the air? Well, we figured that out
in the last video. We're going to be in the air
this long-- 2 s sine of theta divided by g. So the time is going to be
2 s sine of theta over g. So the total distance we're
going to travel, pretty straightforward, rate
times time. It's just the product
of these two things. And we could put all of the
constants out front, so it's a little bit clearer that it's
a function of theta. So we can write that the
distance traveled-- let me do that same green. The distance traveled as a
function of theta is equal to-- I'll do that
in this blue. This s times 2s divided by g
is-- I'll do it in a neutral color actually. This s times 2s divided by g is
2 times s squared over g. So 2s squared over g times
cosine of theta times sine of theta. So now we have a general
function. You give me an angle that I'm
going to shoot something off at and you give me the magnitude
of its velocity, and you give me the acceleration
of gravity. I guess if we were on some
other planet, who knows? And I will tell you
exactly what the horizontal distance is.