youtube algorithm recommended my way into your channel and im liking it so far! very good explanations, animations and content in general. i legit thought you were way more famous, because the quality here is really good! so keep up with the great work, cause i (like many) am really enjoying your videos =)
I solved it a slightly different way assign variables as follows a: angle chosen d: distance to target v: initial velocity of cannon ball g: acceleration due to gravity Then we have the time to target is given by t=d/(v*cos(a)) (since target is at the same height as the cannon we can simplify by cancelling out both the heights of target and cannon) then the height of the cannon at time t is given by v*d*sin(a)/(v*cos(a)) - g*d^2/(2*v^2*cos^2(a)) Which can simplify to d*tan(a)-g*d^2/(2*v^2*cos^2(a)) We want this to be zero so we get d*tan(a)=g*d^2/(2*v^2*cos^2(a)) 2*d*v^2*tan(a)*cos^2(a)=g*d^2 We can cancel out a d on both sides and use tan(a)=sin(a)/cos(a) to get 2v^2*sin(a)*cos(a)=g*d^2 now use sin(2a)=2*sin(a)*cos(a) to get v^2*sin(2a)=g*d sin(2a)=gd/v^2
It's not much more work to solve the following problem: At what angle you must set the cannon to hit a target at a given location (x,y) using the smallest initial velocity possible?
Nice solution. My approach to this problem would be different, since the height of the target is the same as of the cannon, it's going to take a simetric time and trajectory to hit, and the total amount of time is twice the "uphill" time, so, we have 100*sin(θ) - g*t' = 0 100*sin(θ)/g = t' (uphill time) So the time of the whole trajectory is 2*t', which is t = 200*sin(θ)/g But horizontally we have, 1000 = 100*cos(θ)*t 100*200*sin(θ)*cos(θ)/g = 1000 10*2*sin(θ)cos(θ)/g = 1 And, from trigonometry, 2*sin(θ)*cos(θ) = sin(2θ) So, 10sin(2θ)/g = 1 sin(2θ) = 9,8/10 = 0,98 θ is approximately 39° Or 50° Because of the simetric situation we don't need to solve quadratics or find the actual trajectory equation of the launch. Greetings from Brazil, nice channel 👍😃
I have 3 questions. 1) What changes do we have to make for a difference in height of target and cannon? 2) What adjustment would we have to make for a cannon ball of different mass? 3) how do we calculate the magnitude using something like a rubber band setup?
@@suspended3785 surely a cannon with a constant propelling force won't propel a projectile with heavier mass as far as it will propel a projectile with a relatively lighter mass. I think that with the force constant, the velocity of the lighter ball will be higher than that of the heavier ball.
@@barnabasonubi336 This will be the case if you are not neglecting air resistance, drag etc. If those are neglected (like in this video) the mass of ball would not matter on the distance. Only the initial velocity matters. Ps. The range of any projectile is given by R = u²sin(2X)/g where u is the initial velocity and x is the angle in degrees and g is gravitational acceleration.
I learned with the AP Calculus textbook by Larson and Edwards, 10th edition. I still have it, and I've used some problems from it as inspiration for my videos.
That's a great point. The effects of the earth's rotation at this scale are negligible. There are other variables that would have a greater impact, such as wind resistance and drag. My main goal is to demonstrate calculus concepts in a fun way, so I wanted to keep the problem as simple as possible. Thank you for checking out my video!
@@ridwan6695 in my physics undergrad course we were given problems involving hard integrals and derivatives and we always had to show our work, not just mark the right answer.
underated channel
youtube algorithm recommended my way into your channel and im liking it so far! very good explanations, animations and content in general. i legit thought you were way more famous, because the quality here is really good! so keep up with the great work, cause i (like many) am really enjoying your videos =)
Thank you so much for those kind and inspirational words. 😃
Just on time! 7 minutes that cleared a lot of doubts in my mind. Keep making these type of videos, enjoyed and learned a lot. 👏
Thank you! I'm so glad it was helpful!
Wow. This is so much easier with calculus. Im in a calc based physics class, and we aren't using calc. Just algebra. This makes way more sense to me.
Your channel is a hidden gem, the name LearnPlaySolve suits your method of explaining the concepts. Highly impressed!
Keep making such videos. 👍
Thanks a lot! I'm glad you are enjoying it!
the animations are honestly steller, keep up the good work (this channel is gonna blow up for sure!)
Thank you for saying that. 🙂
I solved it a slightly different way
assign variables as follows
a: angle chosen
d: distance to target
v: initial velocity of cannon ball
g: acceleration due to gravity
Then we have the time to target is given by
t=d/(v*cos(a))
(since target is at the same height as the cannon we can simplify by cancelling out both the heights of target and cannon) then the height of the cannon at time t is given by
v*d*sin(a)/(v*cos(a)) - g*d^2/(2*v^2*cos^2(a))
Which can simplify to
d*tan(a)-g*d^2/(2*v^2*cos^2(a))
We want this to be zero so we get
d*tan(a)=g*d^2/(2*v^2*cos^2(a))
2*d*v^2*tan(a)*cos^2(a)=g*d^2
We can cancel out a d on both sides and use tan(a)=sin(a)/cos(a) to get
2v^2*sin(a)*cos(a)=g*d^2
now use sin(2a)=2*sin(a)*cos(a) to get
v^2*sin(2a)=g*d
sin(2a)=gd/v^2
Thank you, trying to calculate the parabola for a shell in my game not irl
thank you
It's not much more work to solve the following problem: At what angle you must set the cannon to hit a target at a given location (x,y) using the smallest initial velocity possible?
Underated channel hope you get a lot of subscribers. Very very informative and clear explanation.
I appreciate that!
Nice solution.
My approach to this problem would be different, since the height of the target is the same as of the cannon, it's going to take a simetric time and trajectory to hit, and the total amount of time is twice the "uphill" time, so, we have 100*sin(θ) - g*t' = 0
100*sin(θ)/g = t' (uphill time)
So the time of the whole trajectory is 2*t', which is t = 200*sin(θ)/g
But horizontally we have, 1000 = 100*cos(θ)*t
100*200*sin(θ)*cos(θ)/g = 1000
10*2*sin(θ)cos(θ)/g = 1
And, from trigonometry, 2*sin(θ)*cos(θ) = sin(2θ)
So, 10sin(2θ)/g = 1
sin(2θ) = 9,8/10 = 0,98
θ is approximately 39°
Or
50°
Because of the simetric situation we don't need to solve quadratics or find the actual trajectory equation of the launch.
Greetings from Brazil, nice channel 👍😃
I like your way of doing that. It's definitely quicker than what I did and it makes sense intuitively. Thanks
@@LearnPlaySolve Thank you for considering my solution friend
I have 3 questions. 1) What changes do we have to make for a difference in height of target and cannon? 2) What adjustment would we have to make for a cannon ball of different mass? 3) how do we calculate the magnitude using something like a rubber band setup?
I don't think the mass of cannon ball matters as the value of gravitation acceleration is constant for all masses
@@suspended3785 surely a cannon with a constant propelling force won't propel a projectile with heavier mass as far as it will propel a projectile with a relatively lighter mass. I think that with the force constant, the velocity of the lighter ball will be higher than that of the heavier ball.
@@barnabasonubi336 This will be the case if you are not neglecting air resistance, drag etc. If those are neglected (like in this video) the mass of ball would not matter on the distance. Only the initial velocity matters.
Ps. The range of any projectile is given by
R = u²sin(2X)/g
where u is the initial velocity and x is the angle in degrees and g is gravitational acceleration.
@@suspended3785 Is there a way I can share a video of this experiment with you, so you see what I'm saying? What I'm using is a catapult setup
@@barnabasonubi336 upload it on UA-cam
If the starting and ending position are at the same level, then complementary angles will reach the same ending position.
Can you suggestions some calculus and physics book teach these things for me. Thank you so much.
I learned with the AP Calculus textbook by Larson and Edwards, 10th edition. I still have it, and I've used some problems from it as inspiration for my videos.
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I'm wondering why in physics never considered the rotation of Earth and all other crazy movements. Is the Earth stationary?
That's a great point. The effects of the earth's rotation at this scale are negligible. There are other variables that would have a greater impact, such as wind resistance and drag. My main goal is to demonstrate calculus concepts in a fun way, so I wanted to keep the problem as simple as possible. Thank you for checking out my video!
Genuinely, Good..
this is the type of question I get in my physics exams lol
so your exames are easy 😅 my exames used to be much more harder.
@@User-jr7vf what type of question u get in ur exams 😦🙁
@@ridwan6695 in my physics undergrad course we were given problems involving hard integrals and derivatives and we always had to show our work, not just mark the right answer.
@@User-jr7vf what makes u think others dont have to show their 'work' too lol. written tests still exist, mcqs havent taken over yet