Launching and landing on different elevations | Two-dimensional motion | Physics | Khan Academy
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- Опубліковано 11 жов 2024
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More complicated example involving launching and landing at different elevations. Created by Sal Khan.
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To solve without quadratic formula:
1. Set final vertical velocity = 0, solve for S using the vertical components, which will give you the distance from the elevation to the maximum. S = ((V^2) - (v^2)) / 2a
2. Use this distance to find the time it takes to reach the maximum. t = 2S / (V + v)
(this is the 1st of the 2 time values we need to add to find the total time)
3. Use the distance from the maximum height to the minimum height to value to find the time, t, it takes to reach the minimum. Since we're only considering vertical components the projectile is basically falling from the maximum, where initial velocity = 0, to the minimum, with an acceleration of 9.8. Thus t = sqrt(2S / a)
4. Add (time it takes to reach max.) + (time it takes from max. to min.) to find the total time.
5. Multiple total time by vertical velocity to find vertical displacement.
Hope this helps :)
My god thank you for Khan Academy. I'm so confused in physics because of projectile motion
Thank you so much.
I think it’s easier to conceptually break this into two parts: decelerating ascent up to 0 m/s vertical velocity, then from max height/0 vertical velocity accelerating down to -16m elevation. The math is easier, too. You get time to go up and add that to time to drop down for total flight time.
Divide the velocity vector into its horizontal & vertical components.
- you use the vertical component/motion to figure out how long the projectile is going to stay in the air
- you use the horizontal component/motion to find how far did it traveled (given how long it stayed in the air)
Great advise Sir Sal!
However, this approach to solve the problem doesn't seem intuitive for me.
I find it labelling parts of the trajectory of the projectile then solving the time in each interval of motion is more intuitive when analyzed part by part using 1D kinematics in the vertical component.
UPDATE:
1. i see that with your solution that signs of every variable matters on the bigger picture. We don't take the motion of the ball as positive direction. It's seeing the motion of the object conventionally, i.e.
up is positive, down is negative
right is positive, left is negative
Wouldn't the projectile gain some elevation before it falls the 16m?
Same... I didn't understand that part.
me neither
I understand that since the initial velocity is positive, the equation is gonna tell us how long it's going to take before the speed reaches 0, i.e max position and then how long it takes to reach that -16 position. But the -16 is calculated from the tip of the cannon. So Isn't the total displacement more than just -16?
Because you're launching it in the air with a vertical speed, making the projectile go over the tip of the cannon, which would make the vertical displacement more than just -16?
I confused lol
ok guys i figured it out haha basically the displacement formula is THE CHANGE in position from initial to final (that's why it's DELTA). The upwards displacement is accounted for in the first part of the equation (vi*t) and then the second part, calculates the downwards motion as it factors in the negative acceleration due to gravity.
I wanted to test this just to be sure so i re-did the previous example (projectile in same elevation) using the displacement formula and it gave me the same answer.
So basically when the initial starting position and the final position are the same, the delta s is equal to 0, because there's no change in position.
Displacement =d1-d2,what happens in between doesnt matter.
Well infact you made a technical misstake. 2 times negative "a" gives a positive 9.8 so yes you correct to say that the positive version of the equation was going to give the positive value of "t". Not that its a fatal misstake just pointing it out :D the equation is still giving the right result.
he was about to make a whole new video on how to find the total displacement lol took him 20 seconds to do
this is exactly what i need. im taking gr 12 physics TT
Just been going through this to revise for exams in a few weeks... I want to cry.
I hate biomechanics.
good question, have my like.
REALLY HELPED. I'm grateful... thank you
the distance between the differing plateaus is unspecified, so it was assumed to negligible
How do we decide the vertical displacement is exactly -16m? Can't the projectile move even higher?
You're calculating the displacement and not the total distance travelled.
When finding delta t at the end, shouldn't vertical velocity be negative as well if gravity is negative?
sir can u plz make a video for vectors
JAZAKALLAH . MAY ALLAH BLESS YOU
if a plane ascending at an angle releases a projectile.. will the projectile have a vertical component of velocity... plz answer
Yes. It's called inertia. The projectile will have the same speed as the plane right before it releases it.
is it possible to reach the same conclusion by factoring that quadratic equation?
(A + B)(B-C) like that?
step one. use trig.
Wait, if we're looking for delta t and have the change in y displacement, angle, and initial velocity, why do we need to do quadratic formula? Is it because we don't have horizontal displacement? Or because it's unlevel?
Because the equation he is using is d_y=vit+1/2at^2 since the t is what you are finding and it is squared he used the quadratic
Because it is unlevel.
Thank you so much
Why that initial velocity is 90sin53? Isn't it 90m/s?
90m/s is the initial magnitude of the velocity, but 90sin53 is the vertical component of the velocity. You need the vertical component to find the time delta.
hi! is this method applicable for projectiles launched at lower height and landed on a higher height?
For my numerator i get -52.616 is my calculator in a different function?
@jasonf20 Yes I thought also about the same thing I was just asking my self same question how can we be sure he landed at -16m We could have landed on the lower level.
I guess it would have been more technically correct if he said and we landed on the upper floor.
Nice video though I learned alot from it.
nice video. What about time in cannon it is not gravity affected isn't it ?
Yes but for the scope of the video, it is unimportant
why dont the units equal seconds? they seem to become 1/sec
where is video number 1? -..-
Süper
JACK JACK!
...Huh?
Pls I used my own calculator and the values aren’t corresponding
Can you repeat each sentence 3 times instead of 2 please, its not very clear
i love you
only 4 comments?!?!?!
how did he derive that formulae tho
Do you have to repeat every line twice
EXACTLY!!!!!! That triggered me soooo much. dont even know why
Love this. helps it sink in.
The correct answer is : 806.39146874175632236674 (m)
first
NANI
you made a mistake, you did +4AC instead of -4AC in the calculator
No he is correct, he was multiplying a negative times a negative so he got a positive value. He could of typed out the actual values and it would be the same answer.
Why do you repeat every word you say
You know there's a thing called editing.
sir can u plz make a video for vectors
Thats literally the very first video in this series.