Projectile Motion: A Vector Calculus Problem

[nyyakko]

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 =)

[Mike-ksqu]

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.

[barnabasonubi]

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?

[happyhomosapien]

Your channel is a hidden gem, the name LearnPlaySolve suits your method of explaining the concepts. Highly impressed!
Keep making such videos. 👍

[pinpoint]

the animations are honestly steller, keep up the good work (this channel is gonna blow up for sure!)

[gabrieldealbuquerque]

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. 👏

[solomonarlanc.]

Underated channel hope you get a lot of subscribers. Very very informative and clear explanation.

[manuelb__r]

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 👍😃

[ChrisShannon]

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?

[boeroeng]

Thank you, trying to calculate the parabola for a shell in my game not irl

thank you

[dungduong]

Can you suggestions some calculus and physics book teach these things for me. Thank you so much.

[Caio_Myguel]

If the starting and ending position are at the same level, then complementary angles will reach the same ending position.

[danielbranscombe]

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



We want this to be zero so we get


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

[icyify]

this is the type of question I get in my physics exams lol

[riyadhnaqashi]

I'm wondering why in physics never considered the rotation of Earth and all other crazy movements. Is the Earth stationary?