Question

A model airplane is shot up from a platform 1 foot above the ground with an initial upward velocity of 56 feet per second. The height of the airplane above the ground after t seconds is given by the equation h=−16t²+56t+1, where h is the height of the airplane in feet and t is the time in seconds after it is launched. What is the maximum height the airplane reaches, to the nearest foot?
A:30ft
B:50ft
C:60ft
D:80ft

Answer 1

\[h(t) = -16t^2+56t+1\]

That's a parabola. The vertex of the parabola is the maximum height the airplane will reach.

Answer 2

You can find the vertex of a parabola in the form \[y = ax^2+bx+c\] by evaluating \[x = -\frac{b}{2a}\]and then evaluating the function at that value of \(x\) to get the corresponding \(y\) value

Answer 3

so x=-56/2(-16)

Answer 4

good so far...

Answer 5

x=7/4 is what i got from that

Answer 6

yes, that's correct. what will the value of \(h(\frac{7}{4})=\)

Answer 7

so we solved for t correct?

Answer 8

\(t=7/4\), yes

now you need to find the actual height at that time.

Answer 9

\[h(\frac{ 7 }{ 4 }) = -16(\frac{ 7 }{ 4 }) + 56(\frac{ 7 }{ 4 }) +1\]

Answer 10

no, the first term has t^2, right?

Answer 11

wait -16(7/4) is squared

Answer 12

snap :-)

Answer 13

only the 7/4 is squared, not the -16

Answer 14

ok so i got 28.5 would the anwser be A?

Answer 15

cause it asks for the nearest foot

Answer 16

i did this wrong... sorry to waste your time thanks for the help anyway

Answer 17

Hmm, looks like you did the arithmetic incorrectly.

\[h(\frac{7}{4}) = -16(\frac{7}{4})^2+56(\frac{7}{4})+1\]\[\qquad=-16(\frac{49}{16})+14*7 + 1\]\[\qquad=-49+98+1\]\[\qquad=50\]

Here's a plot of the height vs. time, with grid lines added to show the exact vertex location.