I don't understand this: An object is launched upward at 64 ft/sec from a platform that is 80 feet high. What is the objects maximum height if the equation of height (h) in terms of time (t) of the object is given by h(t) = -16t2 + 64t + 80?

Question

abb0t · Answer

Find the derivative and find the roots of the function.

bruno102 · Answer

I don't know how to do that.

bruno102 · Answer

Actually, I believe I am understanding it.

bruno102 · Answer

would it be 2, 1232? Or am I totally wrong

phi · Answer

h(t) = -16t2 + 64t + 80
is a parabola

can you find its vertex?  that will be when it is highest

zepdrix · Answer

Hmm I'm not sure how to solve optimization problems like this without taking a derivative D: Hopefully phi can help.

zepdrix · Answer

oh parabola :O hehe

bruno102 · Answer

would it be 1232 feet?

zepdrix · Answer

Factor -16 out of each of the first two terms,
$$\large h(t)=-16(t^2-4t)+80$$
Complete the square on the t's,$$\large h(t)=-16(t^2-4t+4)+64+80$$Which simplifies to,$$\Large h(t)=-16(t-2)^2+144$$
Can you determine the vertex of the parabola now that it's in standard form?

bruno102 · Answer

But what is t?

anonymous · Answer

the first coordianate of the vertex is always $-\frac{b}{2a}$ which as you said, is 2 
the second coordinate is $h(2)$

anonymous · Answer

i.e. plug in 2 where you see $t$ to get the maximum height

anonymous · Answer

$$ h(2) = -16t^2 + 64t + 80$$$$ h(2) = -16\times (2)^2 + 64\times 2 + 80$$

anonymous · Answer

you do not need the derivative, and you do not need the roots
you only need the magic formula for finding the first coordinate of the vertex of a quadratic 
$$y=ax^2+bx+c$$ which is $-\frac{b}{2a}$

anonymous · Answer

you get 
$$h(2)=-16\times 4+128+80=64+80=144$$

bruno102 · Answer

What are the numbers after the equal sign (64+80)?

bruno102 · Answer

what happens to 128 in the equation if the answer is 144?

anonymous · Answer

are you asking how to compute $h(2)$ ?

bruno102 · Answer

I think so.

anonymous · Answer

replace every $t$ in $-16t^2+64t+80$ by a $2$

bruno102 · Answer

Ok I did that and I get $$-16 \times 2^{2} +64 \times 2 + 80$$ right?

anonymous · Answer

yes

bruno102 · Answer

Ohhhh I get it now I was forgetting the negative sign of -64 when I added it to 128 and then  80. So the answer of the question is just 144.

anonymous · Answer

yes

bruno102 · Answer

Ok thank you so much!

anonymous · Answer

yw