Question

A projectile is thrown upward so that its distance above the ground after t seconds is given by the function h(t) = - 16t^2 + 704t. After how many seconds does the projectile take to reach its maximum height?

Answer 1

do you know differential calculus? specifically derivatives?

Answer 2

Derivative: h'(t)=-32t+704
h'(t)=0 then:
t=704/32
t=22

Answer 3

where did you get the -32?

Answer 4

Can't you just find the vertex of the parabola?

Answer 5

derivative of -16t^2..power rule

Answer 6

\[\frac{d}{dx} -16t^2 \implies (2 \times -16)t ^{2-1} \implies -32t\]

Answer 7

then \[\frac{d}{dx} 704t \implies (1 \times 704)t^{1-1} \implies 704t^0 \implies 704\]

Answer 8

so you get \[-32t + 704 = 0\]
\[704 = 32t\]
\[\frac{704}{32} = \frac{32t}{32}\]
\[\frac{704}{32} = t\]

Answer 9

ohhh i think i get it thanks!

Answer 10

you're welcome and glad you got it :)