Question

A model rocket is launched straight upward from the side of a 256-ft cliff. The initial velocity is 96 ft/sec. The height of the rocket h(t) is given by:

h(t) =-16t^2+96t+256

where h(t) is measured in feet and t is the time in seconds. Determine the time at which the rocket is at the maximum height and the maximum height it reaches.

Answer 1

to find the vertex, do -b/2a which gives you 3 (your x value of the vertex). Plug in 3 to get the y value.

x=seconds
y=time

Try factoring the quadratic with 128 and 32 to find what "t" is

Answer 2

that is a parabola, where is the vertex?

16 t^2 - 96 t = -h +256

t^2 - 6 t = (1/16)(-h + 256)

t^2 - 6 t + (6/2)^2 = (1/16)(-h + 256) + 9

(t-3)^2 = (1/16)(-h + 256) +(1/16)(144)

(t-3)^2 = -(1/16)(h-400)

vertex at 3 seconds and h =400 ft

Answer 3

h(t) =-16t^2+96t+256
= -16(t^2-6t-16)
so the roots (h=0) are
(t-8)(t+2)=0, or t=-2, and 8

Becuse this is a parabola, the maximum will occur when t is halfway between, or t=3 (check that).
so the max height occurs when t = 3, solve for h(3).

Answer 4

LOL - three ways to get at that same parabola