Question

Suppose that the height of an object shot straight up is given by h=512t-16t^2 (Here h is in seconds,) Find the max height and the time at which the object hits the ground

Answer 1

h' = 512-32t
t = 512/32 = 16

Answer 2

max height is: 4096

Answer 3

is it being shot from the ground or from somewhere else like a tower?

Answer 4

I assume the group

Answer 5

time when object hits the ground is easy..h=0
512t-16t^2 = 0
t1 = 0
512-16t=0
t = 512/16 = 32

Answer 6

ground

Answer 7

it's not 0, because at 0 it was on the ground, about to get shot into the air

Answer 8

ok

Answer 9

Vertex = highest point: x=-b/2a

second part: Use the quadratic formula. You'll get two solutions for x. The bigger solution for x is the correct one.

Answer 10

h=512t-16t^2, so h = -16t^2 +512t + 0

a = -16
b = 512

t=-b/2a

Once you find your t-value for the vertex, plug it back in to your quadratic equation to find h.

h = -16t^2 +512t + 0

Answer 11

that helped a lot. Thanks a ton