#1

Hello pit, this probably isn't the greatest idea, but I've got this issue.

Here's the equation for the problem:

-16t^2 + 240t + 20.

I need to find out when the rocket reaches maximum height. But I can't remember for the life of me how to do it.

Am I supposed to somehow factor out the equation?

Here's the equation for the problem:

-16t^2 + 240t + 20.

I need to find out when the rocket reaches maximum height. But I can't remember for the life of me how to do it.

Am I supposed to somehow factor out the equation?

#2

There's a math and science hekp thread. Those guys can help the best.

But you need to factorise. Then solve for x.

Id it doesn't factorise nicely stick the coefficients into the quadratic equation.

But you need to factorise. Then solve for x.

Id it doesn't factorise nicely stick the coefficients into the quadratic equation.

#3

Are you considering the ****ing Miracles hypothesis valid?

#4

Factorise like

(xt + z) (pt+s) where x,z and s can be anything you choose that works. Equate each bracket to zero and find 2 values for t.

(xt + z) (pt+s) where x,z and s can be anything you choose that works. Equate each bracket to zero and find 2 values for t.

#5

factor, solve, get rid of negative one?

#6

Is that the equation for position? If so, take the derivative of it and set it equal to 0 and solve for t.

#7

Setting the equation for 0 solves for when it hits the ground. I'm looking through my notes and for whatever reason I never wrote down an example for a problem like this.

EDIT: Found what I needed. Thanks for the help, everyone.

EDIT: Found what I needed. Thanks for the help, everyone.

*Last edited by r0ckth3d34n at Jun 2, 2011,*

#8

Do you know how to take derivatives? Taking the derivative of this

h = -16t^2 + 240t + 20

gets you

dh/dt = -32t + 240

rate of change of height is just velocity, so

v = -32t +240

At the maximum height, the velocity in the y direction will equal 0. So set the equation equal to zero, and solve for t.

I THINK that's how you do it...

h = -16t^2 + 240t + 20

gets you

dh/dt = -32t + 240

rate of change of height is just velocity, so

v = -32t +240

At the maximum height, the velocity in the y direction will equal 0. So set the equation equal to zero, and solve for t.

I THINK that's how you do it...

#9

Do you know how to take derivatives? Taking the derivative of this

h = -16t^2 + 240t + 20

gets you

dh/dt = -32t + 240

rate of change of height is just velocity, so

v = -32t +240

At the maximum height, the velocity in the y direction will equal 0. So set the equation equal to zero, and solve for t.

I THINK that's how you do it...

Yeah, this.

I forgot to make sure I was solving it for the right thing :P

#10

Just for clarification, am I right in assuming that -16 is the acceleration in m/s², 240 is the initial velocity in m/s, and that you've moved a -20 initial distance so that this whole thing = 0? So:

0 = a(t)² + Vi(t) + d?

If I'm right and you're only doing first year high school physics, that's a quadratic bro. Just use the basic quadratic equation of y = ax² + bx + c to substitute your values into the quadratic formula:

Whichever number is positive (you have to do the equation using both the addition and subtraction operations) is your time.

0 = a(t)² + Vi(t) + d?

If I'm right and you're only doing first year high school physics, that's a quadratic bro. Just use the basic quadratic equation of y = ax² + bx + c to substitute your values into the quadratic formula:

Whichever number is positive (you have to do the equation using both the addition and subtraction operations) is your time.

Isn't the quadratic formula taken from the derivative of a quadratic equation? I haven't done derivatives yet so I don't know. But if that's the case then the above way would be easier I think.Do you know how to take derivatives? Taking the derivative of this

(*bah bah bah math*)

At the maximum height, the velocity in the y direction will equal 0. So set the equation equal to zero, and solve for t.

I THINK that's how you do it...

*Last edited by Pat_s1t at Jun 2, 2011,*

#11

Whichever number is positive (you have to do the equation using both the addition and subtraction operations) is your time.Isn't the quadratic formula taken from the derivative of a quadratic equation? I haven't done derivatives yet so I don't know. But if that's the case then the above way would be easier I think.

Derivatives are just differentiation.

It's far easier than plugging into the quadratic equation.

That and it gives more info out.

#12

Take the first derivitave, set it equal to 0, solve for t

#13

I remember hating these problems in 11th and 12th grade Functions, then getting to Calculus and realizing how easy derivatives make them stupid rocket problems...

#14

Haven't done a lick of calculus all through school, taking my first calculus course next year. If it's as easy as you guys say, I will be sad.Derivatives are just differentiation.

It's far easier than plugging into the quadratic equation.

That and it gives more info out.

#15

You guys are making this way too complicated.

Remember that graphing that as a function will give you a parabola that represents the trajectory of the rocket. Essentially, you need to find the vertex (maximum) of that parabola.

So first you find the Axis of symmetry with the formula -b/2a

Then you plug that into the original function to find the vertex.

That's the early algebra method, but you can indeed solve it with a derivative.

Remember that graphing that as a function will give you a parabola that represents the trajectory of the rocket. Essentially, you need to find the vertex (maximum) of that parabola.

So first you find the Axis of symmetry with the formula -b/2a

Then you plug that into the original function to find the vertex.

That's the early algebra method, but you can indeed solve it with a derivative.

*Last edited by TheSPillow at Jun 2, 2011,*