I'm not sure I ever learned that vertex formula (it was 50 years ago, so maybe). I differentiated the equation and solved for v = 0, but I guess that is not the proper approach for pre calculus.
I didn't know the vertex formula either, but I guess it makes sense as you're solving the quadratic formula for when the discriminant is 0 which is actually the same as finding the solution for h(t) = m where m is a constant, actually the maximum value where the straight line is touching the parabola at a single point. So even without knowing m upfront we know that it must be when D is 0, so we are just left with (-b +- 0) / 2a which simplifies to -b / 2a
With using pre-calculus, you would complete the square. I am not going to go through all the work, but when you complete the square for this equation, you will get -16(t-5)^2 + 400. With this equation, you will get the t = 5 sec (the time it reaches its maximum height) and the maximum height is 400ft.
You can do that anyways to prove max is -b/2a. f(x) = ax^2 + bx + c f'(x) = 2ax + b 2ax + b = 0 2ax = -b x = -b/2a So -b/2a is already doing that but it's faster if you just do -b/2a instead of doing it all again.
it *is* quite a bit of fun to play with them. But the formula sure is incorrect (physically): during the first few seconds the rocket experiences the boost due to the water ejected, afterwards only gravity influences it (assuming airless conditions ;-)), so we'd need a piecewise defined formula. Alternatively we can only think about the part after the water runs out, but then we definitely aren't on the ground anymore. That means we'd either have to add a constat term to the formula (the height when acceleration by water stops) or we'd have so solve for a negative height to get when it returns to ground. Oh, and replace t by (t-a) where a is the time acceleration ends. This reminds me ofquite a few "bungee jumping" physics questions - which tend to assume that the bungee cord starts effecting a force the second one jumps (i.e. it has zero length)
This -16t^2 + 160t formula is too linear as a realistic physics model of how NASA determines astrophysics in how a real rocket would perform. It is too complex and beyond the scope of simple physics mathematics. There is also resistance or atmospheric frictional forces in the ozone layers of Earth. It then has a needed fuel expenditure to reach "escape velocity" past the ozone to the vacuum of space final layer to where the rocket can be a satellite of Earth like the moon. Calculations determined radially and proportional to how the moon circles Earth (and it's measured mass to Earth's mass proportional force inverse square distance from the center of Earth point of reference). A rocket will blast off on ground level with a gravitational force different than the gravitational force at its height of fuel consumed. So there will be g(h), gravitational force, comparison at the height it achieved when fuel ran out or was intentionally ended for Earth to ground return fuel needed in a rocket design. During that launch a change of mass was also experienced; but, not due to height achieved but instead a change of mass during measurable time related to fuel burn calculations. It is ideally related to its mass initial - mass final where the mass final is the mass of the rocket plus return to Earth smaller mass of fuel calculations. Even those calculations are models and piece-wise approximate to how rockets from Earth can be satellites to Earth like the moon and to return to Earth with a little more used fuel to at least slow the rocket from too much ozone barrier resistance heat related to return to Earth velocity changes until parachute and eventual total expended fuel to parachute calculations. Some RUclips sites have change of mass rocket physics and some assumed near gravitational force change constant and close enough to zero to be neglected math.
@@lawrencejelsma8118 I somehow don't think a water rocket will achieve escape velocity and leave earths orbit. Or even enters one :-D Or need to think about variations in gravity due to height.
👍 For mentioning t=0, rather than glibly dismissing it. Many maths problems have a similarly surprising result that should only be dismissed with an explanation.
Trajectory doesn't really matter since there's no air resistance. Two objects that start out with the same vertical velocity and experience the same gravity will be at the same height every moment of their flights even if one goes straight up and the other is moving horizontally at a mile per second. They won't start and end at the same place, but we're only talking about height here anyway.
Yes, you would complete the square of the quadratic function and you would get max height at 400ft and it reach that height in 5 sec. The equation is -16(t-5)^2 + 400
I think I speak for everyone who grew up in the United States when I say of course I've played with bottle rockets. That's basically a rite of passage when you turn 5 years old.
Mathematical fun 😊. Physical nonsens 🤔. Answers (acc. to formula given): (a) 256 ft (b) yes (c) 400 ft 🙂👻 P. S. I love this channel! So pls stay with math, don't try physics...😉
Yes, that would be easier. Take the derivative of the function equate to see to get the time it reaches the max height. So you get -32t + 160 = 0 t = 5, f(5) = 400ft
Please make this a series: math basics in real life
i agree
This would be really cool
Agree
I'm not sure I ever learned that vertex formula (it was 50 years ago, so maybe). I differentiated the equation and solved for v = 0, but I guess that is not the proper approach for pre calculus.
I like the calculus way. A simple derivative, making it equal to zero, and you get the 400ft max height as intended.
I didn't know the vertex formula either, but I guess it makes sense as you're solving the quadratic formula for when the discriminant is 0 which is actually the same as finding the solution for h(t) = m where m is a constant, actually the maximum value where the straight line is touching the parabola at a single point.
So even without knowing m upfront we know that it must be when D is 0, so we are just left with (-b +- 0) / 2a which simplifies to -b / 2a
With using pre-calculus, you would complete the square. I am not going to go through all the work, but when you complete the square for this equation, you will get -16(t-5)^2 + 400. With this equation, you will get the t = 5 sec (the time it reaches its maximum height) and the maximum height is 400ft.
You can do that anyways to prove max is -b/2a.
f(x) = ax^2 + bx + c
f'(x) = 2ax + b
2ax + b = 0
2ax = -b
x = -b/2a
So -b/2a is already doing that but it's faster if you just do -b/2a instead of doing it all again.
I want more of this! Simple/intermediate math applications to physics would be great!
it *is* quite a bit of fun to play with them. But the formula sure is incorrect (physically): during the first few seconds the rocket experiences the boost due to the water ejected, afterwards only gravity influences it (assuming airless conditions ;-)), so we'd need a piecewise defined formula. Alternatively we can only think about the part after the water runs out, but then we definitely aren't on the ground anymore. That means we'd either have to add a constat term to the formula (the height when acceleration by water stops) or we'd have so solve for a negative height to get when it returns to ground. Oh, and replace t by (t-a) where a is the time acceleration ends.
This reminds me ofquite a few "bungee jumping" physics questions - which tend to assume that the bungee cord starts effecting a force the second one jumps (i.e. it has zero length)
I agree. One could have used "ejecting a point mass vertically upward in a vacuum" instead.
Yup, this is more like a mortar.
This -16t^2 + 160t formula is too linear as a realistic physics model of how NASA determines astrophysics in how a real rocket would perform. It is too complex and beyond the scope of simple physics mathematics.
There is also resistance or atmospheric frictional forces in the ozone layers of Earth. It then has a needed fuel expenditure to reach "escape velocity" past the ozone to the vacuum of space final layer to where the rocket can be a satellite of Earth like the moon. Calculations determined radially and proportional to how the moon circles Earth (and it's measured mass to Earth's mass proportional force inverse square distance from the center of Earth point of reference).
A rocket will blast off on ground level with a gravitational force different than the gravitational force at its height of fuel consumed. So there will be g(h), gravitational force, comparison at the height it achieved when fuel ran out or was intentionally ended for Earth to ground return fuel needed in a rocket design. During that launch a change of mass was also experienced; but, not due to height achieved but instead a change of mass during measurable time related to fuel burn calculations. It is ideally related to its mass initial - mass final where the mass final is the mass of the rocket plus return to Earth smaller mass of fuel calculations.
Even those calculations are models and piece-wise approximate to how rockets from Earth can be satellites to Earth like the moon and to return to Earth with a little more used fuel to at least slow the rocket from too much ozone barrier resistance heat related to return to Earth velocity changes until parachute and eventual total expended fuel to parachute calculations. Some RUclips sites have change of mass rocket physics and some assumed near gravitational force change constant and close enough to zero to be neglected math.
@@lawrencejelsma8118 I somehow don't think a water rocket will achieve escape velocity and leave earths orbit. Or even enters one :-D Or need to think about variations in gravity due to height.
👍 For mentioning t=0, rather than glibly dismissing it.
Many maths problems have a similarly surprising result that should only be dismissed with an explanation.
What formula is in part (C) I have never seen this before??
I'd rather see HOW to find the maximum. Solve for the slope = 0, using basic math or derivatives.
@@hrayz Bro I know that too but I wanted to know how he was doing it
Are these calculations accounting for trajectory or is it just a random trajectory? I feel like 400 ft high would be an almost 90 degree trajectory.
What does it matter? Height is a vertical dimension with no horizontal component.
Trajectory doesn't really matter since there's no air resistance. Two objects that start out with the same vertical velocity and experience the same gravity will be at the same height every moment of their flights even if one goes straight up and the other is moving horizontally at a mile per second. They won't start and end at the same place, but we're only talking about height here anyway.
Max height half way..0 to 10 half is 5...so max height t=5....find h(5)
I wish I could remember how to do math. Linear algebra was the last thing I studied. That was so long ago and now I struggle with basic arithmetic.
I think I missed a step, as I don't remember the formula max: t=-b/2a.
Yes, you would complete the square of the quadratic function and you would get max height at 400ft and it reach that height in 5 sec. The equation is
-16(t-5)^2 + 400
I think I speak for everyone who grew up in the United States when I say of course I've played with bottle rockets. That's basically a rite of passage when you turn 5 years old.
h should be in meters.
h(t) = - 4.877t^2 + 48.77t
Better now?
Mathematical fun 😊.
Physical nonsens 🤔.
Answers (acc. to formula given):
(a) 256 ft
(b) yes
(c) 400 ft
🙂👻
P. S. I love this channel! So pls stay with math, don't try physics...😉
hmm ive never learnt b/2a formula… instead i complete the square for the max/min point…
Come on, go metric!!
Aye can you a calculus way of doing this?
Yes, that would be easier. Take the derivative of the function equate to see to get the time it reaches the max height. So you get -32t + 160 = 0 t = 5, f(5) = 400ft
@@thomaspaszynski8888 I knew it. That's exactly what I did to be sure.
0:01 Whaa??? When did your hair grow out???
feet :(
Bro💀