S=-16T^2+96T for what time period is the projectile at least 128ft above ground

Answer 1

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

at t=0, h=0

at t=6, h=0 when it hits the ground, it's in the air 6 seconds

-16t^2+96t = 128

16t^2-96t +128=0

divide by 16

t^2 -6t +8=0

factor

(t-4)(t-2)=0

set each factor =0

t=2,4 seconds when it's 128 feet high

from 2 to 4 seconds it's at or above 128 feet

it reaches maximum height in 3 seconds

max height = -16(3^2)+96(3) =288-144=144 feet

for 2 seconds, between t=2 and 4, the projectile is between 128 and 144 feet

velocity= v(t)=-32t+96

v(3)=0

v(2)=32 feet per second

v(4)=-32 feet per second

v(6)=-32(6)+96= 96-192 =-96 feet per second, as it hits the ground

v(0) = 96 seconds as it initially rises

general formula is

h(t)= (a/2)t^2 +vot +ho

vo=initial velocity = 96 ft/sec

ho = initial height = zero

a = acceleration due to gravity = -32 ft/sec^2

another approach is just graph the quadratic

then look for the t values where h=128, the points (2,128) and (4,128)

it's a downward opening parabola with vertex = maximum point = (3,144)

zeros or x intercepts at (0,0) and (6,0)

use a graphing calculator, on line or handheld

Answer 2

First, find the points where the height equals 128 ft by plugging S=128 into the equation.

128=-16t²+96t Set it equal to zero

0=-16t²+96t-128 Solve using the quadratic equation

a=-16

b=96

c=-128

t=2 or t=4

Between 2 and 4 seconds, the projectile is at least 128 ft above ground.