Question

A baseball is thrown up into the air. The ball's height in feet is modeled by the equation h(t) = -16t2 + 32t + 10. (h = height, t = time in seconds)
Find the maximum height of the ball.

Answer 1

if we assume a starting height of 0 that gives us:

-16t^2 +32t; which factors to:

-16t(t+2) when t = 0 or t = 2; we are back on the ground which means it would take half that time to reach its max height; when t=1 right?

Answer 2

the max height whould be; but dbl chk; 42-16 = 26

Answer 3

derive it to be:

-32t + 32 = 0 when t=1

Answer 4

a) .6 seconds to 1.8 seconds
b) .2 seconds to 1.3 seconds
c) .4 seconds to 1.9 seconds
d) .5 seconds to 2.1 seconds

Answer 5

those arent heights, those are times :)

Answer 6

lol sorry i mean
a) a) 10 feet
b) 26 feet
c) 2.3 feet
d) 23 feet

Answer 7

since b = 26; id go with b :)

Answer 8

okay thanks

Answer 9

You need to make the variation table of the equation,
1st, the equation is defined on the interval [0, +inf[
2nd, the derivative is h'(t)= 32 ( -t + 1 )
3rd, its sign : + ....0.....- ( where h'(1)=0)
4th, the variation of h
5th, Hmax= h(1)=26