I remember a very similar problem in the book University Physics With Modern Physics. I used to hate that problem, for some reason I got stuck for a really long time until I finally realized that I should do like you do in this video.
Thankyou sooo muchh!!! Teacher gave projectile problem just like this, apart from it wasnt a motorbike, and we had to calculate the inital velocity. This video is the reason why I was the only one in my whole class to be able to solve that problem!! Thanks for all the videos! Keep up the good work!👍
@@MichelvanBiezen Hey Michel, any chance you got any videos on solving trigonometric equations without graphing please. For example sin2x=-1, and some more complex examples please.
sin(2x) = -1 Therefore 2x = sin^-1(-1) or 2x = 270 degrees ( = 3 pi / 2) Then x = 135 degrees or 3 pi / 4 We have several playlists on trigonometry. Take a look there.
Professor, did you try out the motorcycle stunt for real? Wow, you are a real physicist. And also a very good teacher. You motivate me a lot! Thanks Arin
Hey Michel, I just wanted to let you know that a university in the Netherlands is prompting your online lectures! I'm watching all of them, they are amazingly helpful to us! Thanks :)
it's a trick question! 'Cause Chevy didn't make a 327 in '55, the 327 didn't come out till '63. And it wasn't offered in the Bel Air with a four-barrel carb till '64. However, in 1964, the correct ignition timing would be four degrees before top-dead-center. ;) great movie!
the y equation has v initial and t as unknown same as x equation. Why were you able to use the v initial for the x equation but not for the y equation?
I could not understand that how we get x and y times are equal? for y, delta y is zero okey but at this point range is longer than 20. Isn't y still a positive value when x is 20 as motorcycle is still in the air? thank you for your great videos it helps me so much
The time spent in the air is the same when the object lands. The time for the object to go up and back down MUST be the same as the time it takes for the object to reach the distance.
Can you please help me with this problem ?? * In what corner of launch to the horizontal the projectile reaches 10times greater height than the maximum horizontal distance, in which falls to the ground ?
how would you approach this problem? A stone is thrown at an angle of 30˚ above the horizontal from the top edge of a cliff with an initial speed of 12 m/s. A stopwatch measures the stone's trajectory time from top of cliff to bottom to be 5.6 s. What is the height of the cliff? (Assume air resistance is negligible.)
Fester MP The time in the air (5.6 sec) is determined by the vertical component only. Thus find v initial in the y direction and plug everything into the equation of kinematics: Y = Yo + Voy * t + (1/2) g * t^2
thank you. I did this however, what makes me not confident about my answer is my initial v is the same v initial from when it was thrown, on the y-direction. so what my calculations states is, that if you throw an object with an initial speed toward the air, then the final speed of the ball, when returned to your hand, will be the same as from when it left your hand. only in the negative direction. is this true.
+Lakeisha Shakes Sure can. However rather than trying to remember certain equations that don't always work (depending on the situation), I rather teach the student how to work with the basic equations and develop the answers that way. They gain better understanding of the material that way.
Couldn't you have just used the Range equation and solve for u (the initial velocity) That is u = Square Root( 20 X 9.81/ sin 30deg) = 19.8m/s You can use the Range equation if the take off and landing heights are the same.
Fester, In projectile problems it is assumed that the acceleration is constant. (the acceleration due to gravity) There is no acceleration in the x direction. If there is, then you have to apply the equations of kinematics in both directions. If you have a particular problem in mind in may be easier to let me know what the problem is.
ok so a car is at rest and then it accelerates at 2.0 m/s^2 on a inclined road of 5.5 degrees.what are the distance for the x coordinate and y coordinate.? so what I was confused about was finding the components of acceleration for ax and ay. what I did was find ax with acos5.5 and ay with asin5.5
Fester MP Find the distance as a function of time along the incline first. d = (1/2) a t^2 Then find the x and y component by multiplying d with the cos 5.5 degrees and then with the sin 5.5 degrees
Sometimes I think we should solve for (x+1) instead of x so that the characters in these problems don't clip the edge of the cliffs or walls they're trying to hurdle lol
You just got me unstuck on my engineering problem and probably saved me hours of my life. Thank you so much for the video!
You are welcome!
Engineering problem?wtf I'm just in high school grade-9 and we need to study this sh*t
Hello, I am a college student dying from physics and I just wanna say that you saved my life.
I remember a very similar problem in the book University Physics With Modern Physics.
I used to hate that problem, for some reason I got stuck for a really long time until I finally realized that I should do like you do in this video.
So great! I come to your channel for physics, chemistry and calculus! Awesome videos nothing but respect!
I had my answer 18,9 m/s, I want my 1 m/s back. Thank you so much.
Thankyou sooo muchh!!! Teacher gave projectile problem just like this, apart from it wasnt a motorbike, and we had to calculate the inital velocity. This video is the reason why I was the only one in my whole class to be able to solve that problem!! Thanks for all the videos! Keep up the good work!👍
You're welcome!
@@MichelvanBiezen Hey Michel, any chance you got any videos on solving trigonometric equations without graphing please. For example sin2x=-1, and some more complex examples please.
sin(2x) = -1 Therefore 2x = sin^-1(-1) or 2x = 270 degrees ( = 3 pi / 2) Then x = 135 degrees or 3 pi / 4 We have several playlists on trigonometry. Take a look there.
@@MichelvanBiezen thanks so much!! ,I'll have a look 😁
Professor, did you try out the motorcycle stunt for real?
Wow, you are a real physicist. And also a very good teacher.
You motivate me a lot!
Thanks
Arin
+Arindam Bortamuly
I did used to ride motorcycles, but I was not that crazy to try a stunt like that. I will leave that to others......
+Michel van Biezen loool
i was struggling for so long till i came across this video, thank you so much!
Glad you found us.
Wow! I remember Evil Knlevel! My dad worked for a chain mfg. plant....Evil visited the plant and endorsed the quality of the chain! 😊
Wow, a small world.
Sir you are the GOAT of all GOATS.
Thank you. Glad you find our videos helpful.
You are a incredible teacher, thank you so much for all videos.
Hey Michel, I just wanted to let you know that a university in the Netherlands is prompting your online lectures! I'm watching all of them, they are amazingly helpful to us! Thanks :)
Thank you for the feedback. Where is that university? (Ik was geboren in een klein dorpje in Belgie. )
@@MichelvanBiezen Ahha! Aan uw naam te zien dacht ik al dat u waarschijnlijk uit de Benelux komt. Dit betreft de TU eindhoven, waar bent u professor?
Op een universiteit in Los Angeles. (Loyola Marymount University).
it's simple if u use R=square of u × sin2@ upon g
top quality video Mr Michel. I can make use of these examples immediately
How to solve if angle not given...and motorist jumps from certain heigth?
👍🏼👍🏼Professor I been learning a ton from you ! Thank you✌🏼✌🏼
Way freaking awesome. You are a math wizzard.
Glad it helped!
thanks for uploading. u really deserve appreciation sir.
Since g = - 9.8 m/sec^2
Professor,I have a question. lets say we had 40 m width of canyon instead of 20 m. Then does V initial would be the same amount?
Can you solve this using the range equation? and setting 20 meters as your range?
Yes, you can.
also you can solve it by one move using
x= v(square) sinθ / g
Thank you so much, you have saved my life!
What would the correct ignition timing be on a 1955 Bel Air Chevrolet, with a 327 cubic-inch engine and a four-barrel carburetor?
it's a trick question! 'Cause Chevy didn't make a 327 in '55, the 327 didn't come out till '63. And it wasn't offered in the Bel Air with a four-barrel carb till '64. However, in 1964, the correct ignition timing would be four degrees before top-dead-center. ;)
great movie!
what would the initial speed need to be, if the angle is 0, to make the 20 meter gap?
At a zero degree angle you cannot clear any gap. (the speed would have to be infinite)
How does the 20 feet apply to this? I'm trying to do this same equation with 10 feet
Then replace the "20" with "10" in the second equation for the horizontal component. Everything else will remain the same.
Do you have a spreedsheet where I can put the values, weight of vehicle, distance of jump, speed, etc to calculate it?
thankyou sir, it so much help me for doing my task😆
Glad to hear that
the y equation has v initial and t as unknown same as x equation. Why were you able to use the v initial for the x equation but not for the y equation?
We found both the x-component of the initial velocity and the y-component of the initial velocity and used it in both the x and y equations.
@@MichelvanBiezenOh I misread things. Thank you!
I don't know how, but I got exactly 14 as the result.
Sometimes physics is tricky...
thank you so much for these videos.
I could not understand that how we get x and y times are equal? for y, delta y is zero okey but at this point range is longer than 20. Isn't y still a positive value when x is 20 as motorcycle is still in the air? thank you for your great videos it helps me so much
The time spent in the air is the same when the object lands. The time for the object to go up and back down MUST be the same as the time it takes for the object to reach the distance.
thank you so much sir:)) @@MichelvanBiezen
Can you please help me with this problem ??
* In what corner of launch to the horizontal the projectile reaches 10times greater height than the maximum horizontal distance, in which falls to the ground ?
why is time in the air in x direction equal to time in the air in y direction?
Sir finding velocity of decline in different high of start point and final point
how would you approach this problem?
A stone is thrown at an angle of 30˚ above the horizontal from the top edge of a cliff with an initial speed of 12 m/s. A stopwatch measures the stone's trajectory time from top of cliff to bottom to be 5.6 s. What is the height of the cliff? (Assume air resistance is negligible.)
Fester MP
The time in the air (5.6 sec) is determined by the vertical component only. Thus find v initial in the y direction and plug everything into the equation of kinematics:
Y = Yo + Voy * t + (1/2) g * t^2
thank you. I did this however, what makes me not confident about my answer is my initial v is the same v initial from when it was thrown, on the y-direction. so what my calculations states is, that if you throw an object with an initial speed toward the air, then the final speed of the ball, when returned to your hand, will be the same as from when it left your hand. only in the negative direction. is this true.
yes,that's true
Why do you have a + before the 1/2gt^2 and then you use minus?
i just want to say thankyou very very much
cant you just use the horizontal range equation (delta x=Vi^2 x sin2theda/g) and rearrange for Vi
+Lakeisha Shakes because you already have the range and angle + you already know what the force of gravity is
+Lakeisha Shakes
Sure can. However rather than trying to remember certain equations that don't always work (depending on the situation), I rather teach the student how to work with the basic equations and develop the answers that way. They gain better understanding of the material that way.
This one definitely is brilliant!!!!
Why is the displacement zero?
0=5.3-2100/V^2
+Justin Meram
In what direction?
+Michel van Biezen
Oh my bad, you mentioned it was the initial height
Sir when landing plane make an angle how will u find sir
Find the inverse tangent of the ration of the final horizontal velocity and vertical velocity components.
Thank You ,, Thank You ,, Thank You ,,"))
U are So Perfect Sir ..'')))))
Couldn't you have just used the Range equation and solve for u (the initial velocity)
That is u = Square Root( 20 X 9.81/ sin 30deg) = 19.8m/s
You can use the Range equation if the take off and landing heights are the same.
I see this was posted two years ago lol but that is actually incorrect. Your answer would not give you 19.8m/s but 28m/s
Kai Evans Hi The above calculation gives 19.8m/s
@@rumpelforeskin8461 you brains it bruh
Very interesting that 15º just so happens to produce a distance of almost exactly the speed value in m.
I can assure you that it is purely coincidence, interesting, but coincidental.
what if you only have the acceleration. how would you find ax and ay
Fester,
In projectile problems it is assumed that the acceleration is constant. (the acceleration due to gravity)
There is no acceleration in the x direction.
If there is, then you have to apply the equations of kinematics in both directions.
If you have a particular problem in mind in may be easier to let me know what the problem is.
ok so a car is at rest and then it accelerates at 2.0 m/s^2 on a inclined road of 5.5 degrees.what are the distance for the x coordinate and y coordinate.? so what I was confused about was finding the components of acceleration for ax and ay. what I did was find ax with acos5.5 and ay with asin5.5
Fester MP
Find the distance as a function of time along the incline first.
d = (1/2) a t^2
Then find the x and y component by multiplying d with the cos 5.5 degrees and then with the sin 5.5 degrees
thank you!!!!!!!!
I must of did a Ditch Day. I am still in pain because of it ...
where he got the 4.9 from? it it 9.8?
the constant of the acceleration term = (1/2)g = (1/2) * 9.8 m/sec^2
Sometimes I think we should solve for (x+1) instead of x so that the characters in these problems don't clip the edge of the cliffs or walls they're trying to hurdle lol
How does -5.3=-2100/v^2 become v^2=-2100/-5.3
Multiply both sides by v^2. Then divide both sides by -5.3
Thanks, i understand now
That's a lot of math.......... Maybe that's why Evel made jumps by the seat of his pants?? Haha
tnx it helped me alot
About 48 mph should do it
Actually you have 😃
No, I am not that crazy.
this is killing me but ty
Don't give up. Keep at it. The technique is similar to many other projectile motion problems.