Finding the restricted domain of a time/distance problem.

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January 12th 2011, 03:45 PM
Apathy

Finding the restricted domain of a time/distance problem.

The question is: After a football is punted its height, $h$ , in meters above the ground at $t$ seconds can be modelled by $h(t)=-4.9t^2+21t +0.45$ . Determine the restricted domain of this model.

I was also asked to find the maximum height and what time it reached that height and had no difficulty.

I have no idea at all how to find the restricted domain.
January 12th 2011, 03:51 PM
dwsmith

What are the t intercepts? (Time can't be negative)

Max height: set the first derivative equal to 0 and solve for t.
January 12th 2011, 03:57 PM
Apathy

I've already solved the max height and solved the time it reached the height. The T intercept isn't indicated, is that what I am looking for to find the restricted domain? Should I do y=mx+b and find it?
January 12th 2011, 03:58 PM
HallsofIvy

That formula doesn't apply until the ball is kicked (t= 0) and does not apply after the ball hits the ground, h(t)= 0. dwsmith said "t-intercepts" but I think he mean "h-intercepts", where h= 0. However, the ball is kicked from a height of .45 meters, not 0, so only the positive h-intercept is valid.
January 12th 2011, 03:59 PM
dwsmith

You need to find where the parabola crosses the t-axis.
January 12th 2011, 04:07 PM
Apathy

the answer in the book is 0<t<or=4.3. The 0 is logical (although shouldn't I prove it somehow?) but I don't know where the 4.3 comes from.
January 12th 2011, 04:13 PM
dwsmith

Quote:

Originally Posted by Apathy

the answer in the book is 0<t<or=4.3. The 0 is logical (although shouldn't I prove it somehow?) but I don't know where the 4.3 comes from.

$h(t)=-4.9t^2+21t +0.45=-4.9(t-4.307036...)(t+.02132248....)=0\Rightarrow t=-.021, \ 4.307$

Time starts at 0 so only one t value is valid.
January 12th 2011, 04:27 PM
Apathy

Quote:

Originally Posted by dwsmith

$h(t)=-4.9^2+21t +0.45=-4.9(t-4.307036...)(t+.02132248....)=0\Rightarrow t=-.021, \ 4.307$

Time starts at 0 so only one t value is valid.

Sorry It's actually $h(t)=-4.9t^2+21t +0.45$

SO I should sub in 0... that would get me 2 answers and I just reject the one below 0. Does that sound good?
January 12th 2011, 04:28 PM
dwsmith

That was a typo. The factored form is still correct.
January 12th 2011, 04:29 PM
dwsmith

Quote:

Originally Posted by Apathy

Sorry It's actually $h(t)=-4.9t^2+21t +0.45$

SO I should sub in 0... that would get me 2 answers and I just reject the one below 0. Does that sound good?

Sub 0 into where?

Set h(t) = 0
January 12th 2011, 04:40 PM
Apathy

Wait you just factored out -4.9? That would still leave a t^2... the only way I could think of factoring it otherwise would be product/sum method but that would be very hard with decimals like this...
January 12th 2011, 06:49 PM
mr fantastic

Quote:

Originally Posted by Apathy

Wait you just factored out -4.9? That would still leave a t^2... the only way I could think of factoring it otherwise would be product/sum method but that would be very hard with decimals like this...

You need to solve a quadratic equation. Have you been taught how to do so eg. using quadratic formula?