Math Help Please

[-GS-]

Sorry guys, i know this isnt a homework forum or anythign, but i really need help with this thing.

Aite well just saying that i had an equation that was

h = height
t = time

h=-4.9t²+22.8t

1)How would i go about finding when the ball was at its maximum height?
2)And how would i go about finding out how long it was in the air for?

[KustmAce]

Fuck math...

[RSX-S777]

Wouldnt you need the velocity of the ball and/or the trajectory angle to use that equation?

[Raz_Kaz]

You should learn derivables...
Anyways you acheive maximum height at 2.33 seconds. Replace that time for "t" and you get 26.52. Now your maximum height is during half the travel, so the duration in air equals to 2 times max height time which is 4.66

Hope you get your answer in time

[freakray]

Originally Posted by RSX-S777

Wouldnt you need the velocity of the ball and/or the trajectory angle to use that equation?

No.

You already know that the ball deccelerates and accelerates at gravity's constant for acceleration.
Given that you should be able to solve that equation easily enough.

You have to use other velocity equations in conjunction with the one given to resolve it though.

[-GS-]

Heh im with you on this one KustmAce, i hate math.

oh yeh and Raz_Kaz thanks for you answer bud, but may i ask how you go to that, so i may use it for further questions and stuff, like i just found out a way to do it myself by completing the square, but would you be able to tell me your way.

Thanks for the help buddy

[Partizan]

Ok I always went liek this,
h=-4.9t²+22.8t
let h=y and t=x
y=-4.9x²+22.8x (Now common factor)
y=x(-4.9x+22.8) (Now find the roots)
x=0 x=-4.9+22.8
And I can't remeber how the rest goes but basically you want to find the vertex of the parabola and the y value will be the optimal height and the x value will be the time it took to get there. (The total time in the air will be double the x value of the vertex since it is a midpoint). Sorry I couldn't help more but it's been a while and I sucked at it. Good luck.

[Raz_Kaz]

Well GS, I'm not good at explaining but i'll give it a shot...
what your doing is derivables (something we're learning in grade 2 calculus here in Ontario)
Anyways here it goes....
h'=2(-4.9)+22.8, which is the drivable of your previous euation.
Complete it and you have h'=-9.8+22.8
Now you get a max or min when the derivable equals 0 so....0=-9.8t=22.8
Bring the 22.8 on the other side and you get -22.8=-9.8t
t=-22.8/-9.8....which t=2.33
Now you have the time when your going to have a max or min, so replace the value of "t" in the orginal equation
h=-4.9(2.33squared)+22.8(2.33)...And you get the value of h=26.52
Now the max or min of a projectile is in the middle, since you have the time where you reach that maximum or minimum, double it and you have the duration of the trajectory...


Hope thats clear enough

[dayna240sx]

graphing calculator. The best $120 you'll ever spend.

[Raz_Kaz]

/\ Cheap!!! :P...even though it works, what will you do when you don't have it? I don't like math but still...my Calculus teacher is a calculator, how many here can the the third root of sin 45 in their heads??? He can

[zebrathree]

Too many numbers...head exploding.....

[-GS-]

ahhhhhh sick thanks alot both of you guys, appreciate your help. This is the way i found out how to do it like 20 minutes ago, well i knew it before, but i didnt remember it all.

This is wat i did step by step,

h=-4.9t2+22.8t

Then i went ahead and completed the square which got me

h= -4.9 (t-4.65 + 5.405625 - 5.405625)
h= -4.9 (t-2.325)² + 26.4875625

and one of my friends told me that the height would be 26.4875625

[RSX-S777]

At what height will the remnants of Zebrathree's head decelerate and start the descent back toward earth?

[zebrathree]

I NEVER decelerate! How dare you imply that I do!

[Raz_Kaz]

LOL, let me work on that...but im going to need his head weight, initial velocity and more info
But thats getting into physics not calculus




On a side note, GS, learn the damn derivables, so much easier, trust me...