Skip to main content
gamedev.net
🔒 Locked

Trajectory Function -- max height

Started by Halsafar Oct 17, 2005 at 6:04 PM 17 replies 1.3k views
Original Post
Halsafar
Halsafar
Okay I got this function: h = -5t^2 + 140t h = height above the ground in meters t = time in seconds Find the following: a) time required to reach max height b) the max height c) the height above the ground after 7.5seconds - I'm not sure where to start on a) - I assume if I solve a) then b) will fall into place - c) can be solved by plugging in 7.5 for t and solving for h Any assistance?
Zakwayda
Zakwayda
Quote:
Any assistance?
Obligatory question: is it homework?
smitty1276
smitty1276
Do you want the algebra way or the calculus way? And like the other guy said, is this homework?
timw
timw
try to ask yourself the question:

what happes when an object reaches it's maximum height? something special happens to the velocity in the y direction. I think I given enough hint.
Halsafar
Halsafar
Yes this is homework, its part of today's lesson. Worth 0 marks, just an exercise. Whether I do it or not just resolves my understanding of it or not. The teacher doesn't care either way, he won't be reviewing if I did this or not. I do however expect a similar question on the upcoming test which is why I am asking for help.
Currently I do not understand how to get anywhere on it.


The Algebra way please.

When an object reaches its max height its y velocity will start to decrease obviously. But I cannot sit here and punch in random numbers until I get to the point where its y velocity starts to decrease from the highest number I can find.

Its a quadratic, but I am not sure which method to go about solving it at a maximum. Would it be the highest of the roots? Do I use the determinant in anyway? How can I solve the roots when I don't have a value for h.
nilkn
nilkn
Quote:
Original post by Halsafar
When an object reaches its max height its y velocity will start to decrease obviously.


Yes, you are correct. However, I suspect timw wants something else: what happens to the curve, not after the maximum point, but at the maximum point? At this precise point the curve will possess a unique property possessed nowhere else on the curve.
Halsafar
Halsafar
That max value would be the vertex?
Giving us our axis of symetry.

I don't see how that relates tho.
Thermodynamics
Thermodynamics
They are showing you the calculus version and telling you to use a derivative. I would guess that you have not studied that yet.

Let's look at the algebra way to do this.

First hint: Parabolas are nice in that they are symetric. Even in time.

Second hint: If you know when it hits the ground on the other side... when is it at its peak?

I think I have said enough for now.

Edit: Three posts in a row starting with "Unfortunately." Had to change it.
[s]I am a signature virus. Please add me to your signature so that I may multiply.[/s]I am a signature anti-virus. Please use me to remove your signature virus.
Halsafar
Halsafar
I got nothing from that...

h = -5t^2 + 140t
-5t^2 + 140t - h = 0

I cannot find the zero's of this equation since h is undefined.
So I do not know when its crossing the x-axis so I cannot determine when it hits the ground.

maybe -- Determinant

D = 140^2 - 4(-5)(h)
D = 19600 + 20h
20h = 19600
h = 980

Now can I continue using this equation?
-5t^2 + 140t - 980 = 0
Alrecenk
Alrecenk
Well, since you don't seem to know the calculus way, I think the simplest way to solve a problem like this is the physics way (You didn't say what class this was for so for all I know it might be physics). Anyways, in physics you would have access to equations for problems like this. One of which is: h=Vi*t+0.5*A*t^2 . From that it's easy to find initial velocity and acceleration. As was hinted, at the time of the max height velocity = 0. You can find time by using t = (change in V)/(average A) . To find b and c you just substitute t in.
Halsafar
Halsafar
I'd love to solve it using physics since I have some practice in that but I doubt the teacher would be very happy with that.

The course is MathB30 covers Matrices, Complex Numbers, Combination and Permutations, Polynomials, Quadratics. A few more things as well but I don't have the listing near me right now.


My last post had some work done on it. I think I may be on the right track.
Alrecenk
Alrecenk
You could solve for when h=0. When you have the 2 zeros you can find the time of the max height becuase the parabola is symmetric.
nilkn
nilkn
Sorry, the above post was mine. It automatically logged out on me.

Since I can't edit my previous post to add in some more information, I'll just put it here.

A 'zero' of a quadratic function Q(x) is a value of x such that Q(x) = 0. In other words, the zeroes of a quadratic are all the solutions such that the resultant point lies on the x-axis, or equivalently with a y-coordinate of 0, hence the name.

Since Q(x) is your y-coordinate under a disguise, you can find the zeroes by solving Q(x) = 0. In your case, you have the function h(t) = -5t2 + 140t, and so you can determine its zeroes by solving -5t2 + 140t = 0.
Thermodynamics
Thermodynamics
h is defined. _by the equation_ When h is 0 it is resting on the ground. in other words h=0... use the quadratic to solve for t. divide by 2 to get the peak. use that t in the original formula to find the max height. plug 7.5 in the formula for part c.

done
[s]I am a signature virus. Please add me to your signature so that I may multiply.[/s]I am a signature anti-virus. Please use me to remove your signature virus.
Halsafar
Halsafar
Where did divide by 2 come from?


When h = 0 it is on the ground... boy do I feel stupid.
I shoulda caught that.

[Edited by - Halsafar on October 18, 2005 1:12:49 AM]
timw
timw
here are 3 ways to solve

1)
do you know the familiar equation from algebra for the vertex of a parabola?

ax^2 + bx + c = y

x = -b/2*a gives you the x coordiante of the vertex point.

2)
if this was a more physical problem you could formulate it it terms of velocity and note that the velocity in the y direction must be zero at the vertex of the hyperbolic path, formulate an expression for velocity in terms of t, solve for t when v is zero.

3) differientiate the function with respect to t and set it to zero

dH/dT = 0

and slove for t, this t is the t such that the height is maxemized, this is the calculus derivation of -b/2a. the vertex obviously lies at the point where the tangent line has zero slope.

also, someone mentioned division by two, that'd work too.
Tim
nilkn
nilkn
Quote:
Original post by Halsafar
Where did divide by 2 come from?


When h = 0 it is on the ground... boy do I feel stupid.
I shoulda caught that.


The two zeroes of the function are 0 and 28. Since you know the curve will be symmetric, you can reason out that the vertex will occur for the t half-way between 0 and 28, when t = 14.

Plugging t = 14 into your original equation gives you the height at that point.

Topic Locked

This topic has been locked by a moderator. New replies are not allowed.

Sign in to reply to this topic.