- 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
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)
- 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)
- 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)
- how do I calculate the time an object is in the air given the velocity and the angle at which it was launched(1 vote)
- First you find the initial velocity's vertical component which is velocity*sin(angle). Then use that initial upwards velocity along with delta y=0 and the acceleration (-9.8 m/s^2 on earth) to calculate the time it spends in the air.
Did that help?(2 votes)
- What Sal said in this video all makes sense. However, in the previous videos he uses another formula to solve for displacement in the x direction (D= Velocity initial x time + 1/2 x acceleration x sqrt(time)) How come BOTH formulas work? Am I missing something?(1 vote)
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.