Question

i tried and tried, can't figure it out...what is the maximum height of a cliff diver if he jumps from 48 feet at 8ft/sec?

Answer 1

i'm supposed to use this equation h=-16t^2 + vt + s
h : the vertical component of the diver’s motion
v : The initial upward velocity
s : The height of the ledge above the water
t : the time in seconds

Answer 2

help please, its due tomorrow...

Answer 3

lol

Answer 4

its not funny...

Answer 5

in this case

v = 8 and s = 48

so

h=-16t^2 + vt + s

turns into

h=-16t^2 + 8t + 48

Answer 6

h=-16t^2 + 8t + 48 graphs out a parabola that opens downward

The vertex is the point where the max height is reached

Use this formula to find the vertex

t = -b/(2a)

t = -8/(2*(-16)) ... plug in a = -16 and b = 8

t = -8/(-32)

t = 1/4

t = 0.25

This means that at t = 0.25 seconds, the person is at max height

Answer 7

ok, then i plug it in, right?

Answer 8

its 49?

Answer 9

Plug in t = 0.25 to get

h=-16t^2 + 8t + 48

h=-16(0.25)^2 + 8(0.25) + 48

h=-16(0.0625) + 8(0.25) + 48

h=-1 + 2 + 48

h = 49

So the person is at the max height of 49 ft (at the vertex) when the time is t = 0.25 seconds

Answer 10

thanks!

Answer 11

np