This physics video tutorial on projectile motion explains how to calculate the angle at which a projectile launched from the ground will have a maximum range using calculus.
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@jjeeqq Says:
Add curvature of surface (ball) and place a target on that curvature, then calculate angle to shoot that target with best efficient. This formula doesn't work on curvature.
@KodxQ8 Says:
I used to watch these videos for education in high school/university days. But now as a physics teacher, I watch them for fun.
@antoniovivaldi9999 Says:
Thank you so much!
@thenomad5652 Says:
I was hell confused cause that I studied that slope part long ago now I get know how graphically slope is max at 0
@8Space Says:
For anyone who wants to know how to find the equation for the range:
The range will be equal to the initial velocity in the horizontal direction multiplied to the time if there is no drag.
R = Vix(t)
Now to solve for time, imagine you are thrown vertically and you are deccelerating because of the gravity by for example 9.81 m/s² that means each second you lost 9.81 m/s of velocity. So if you were thrown vertically by 9.81 m/s in 1 second your velocity will become 0 then you will accelerate downward due to gravity until you return to your original position. The time it took you to lose the speed you had upward, will be the same time it will take you to return to the ground.
so t = 2(Viy/g)
Now R will be equal to
= Vix(2)(Viy/g)
The horizontal component can be solved using trigonometry. Vix = Vi cos(theta)
Viy = Vi sin(theta)
So
Vix(Viy)
= Vi²(sin(theta))(cos(theta))
Now R becomes:
= [Vi²•2•(sin(θ)•cos(θ))]/g
2sin(θ)cos(θ) = sin(2θ)
from trig angle addition identity:
sin(A + B) = sin(A)cos(B) + sin(B)cos(A)
sin(2A) = sin(A + A)
Now for the final result
R = [(Vi)²(sin(2θ))]/g
I'm not sure if I explained that well.
@SamDomaloy Says:
WTF??!
@kaiser7637 Says:
I looked everywhere but I can't find an explanation as to why the range equation is the way it is. I can't seem to factor it in a way that gets gravity (g) on the bottom
@baconman132 Says:
so is this just optimization? been learning that lately, and it feels familiar
@cianconway2540 Says:
Is range being v^2sin(2 theta) all over g a given or how do I calculate it if asked to show the max range of a particle in terms of v and theta
@showtimebaby7 Says:
SELAM OLSUN
@NotAMimic02 Says:
Hi
@redrevyol Says:
You can also use sin(2*theta) = 1 because one is the maximum value for sine.
@TheEngineerBlueprint Says:
One question closer to acing that exam in 22 days. Thank you!
@deltaepsilonphi Says:
The best tutor!
@ALPHIAMONTALES Says:
Wow cool ahhhhhhhhhhhhhh 😏
@NoctisTheopilus Says:
Bro I don't want to launch my Balls!
@sussushi Says:
genuinely, quite literally, the realest munda on the planet
@rcasewst Says:
Did not account for drag
@guidoreuter6032 Says:
Assuming the firing occurs on a planet void of atmosphere
@bagda4667 Says:
45degree 😂😂😂😂😂😂
@gui-zx3di Says:
9/2v**2 *0?? Is it not wrong to multiply it knowing that this side is 0?
@SarrouTube Says:
Amazing video !! The theory behind the experiment ! Keep uploading !!
@AceAdora Says:
could you make a video about all the set theory proofs and practice problems ? thanks !
@AhShtHereWeeGoAgain Says:
Thanks a lot 😊
@victoryfirst06 Says:
Do you plan to make a video about maximum range angle from non-leveled surfaces as well? I would be very interested to see that one.
@ZurerBW Says:
Two days later is my exam and projectile motion is in the syllabusm, What a coincidence
@thatomofolo452 Says:
Goat
@conspis-gq3ug Says:
when is face reveal
@victorpaulsacyabon7664 Says:
can you explain the trigonometric identities super hexagon?
@Crissss-b5z Says:
Sir you are good 👍 teacher I speak broken English but I can understand what you say from 🇳🇵
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