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YAHT:Is this really hard?

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Lifer
Yet Another homework help thread

If an object is launched straight up into the air from a starting height of "h" feet, then the height of the object after "t" seconds is approximately H=16t^2+vt+h where v is the inital velocity of the object. Find the starting hight and the initial velocity of an object that attains a maximum height of 412 feet five seconds after being launched.

Alright...

I know have the points 5 and 412 which leads me too... 412=-16t^2+v(5)+h

Apparently I am suppose to use f(x)=ax^2+bx+c and -b/2a and f(-b/2a) somehow this is suppose to give me -1/32 somehow.... but i dont remember how exactly?

I also know the velocity is distance over time so 412/5=82.4 is velocity, but something doesn't seem right about that.... Any ideas?

There's also another question h(x)=2x^2+8x-15 and g(x)=2x^2-23 find a linear function f so that h=g(f(x))
 
Originally posted by: newParadigm
Some one help this poor chap, I haven't taken Physics or Calc or w/e this ****** is yet...
Click to expand...

Actually, this is math 103 😉 I'll goto class in about two hours and the teacher will explain this in less than a minute... 😉
 
I talked to my math whiz friend and this is what he told me

[16:10] him: i don't have units so initial velocity is 160
[16:10]him:and the equation is wrong
[16:10] him:suposed to be -16t^2
[16:10] him:160 should be the answer
[16:11] me: how do you know this?
[16:11]him: because it's the equation of position in a gravitational force field
[16:11]him: y position
[16:11] [16:11] him: and -16 is the acceleration in the y direction in feet persecond squared

Hope this helps 🙂
 
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