An object is launched into the air from a ledge of 16 feet off the ground with an initial vertical velocity of 96 feet per second. h = -16t^2 - 96t + 16. Answer the questions below.

Question

At what time is object at a height of 120 feet off the ground?

What is the object's maximum height?

Answer 1

NO! It is y = -16t^2 + 96t + 16

the object is moving upwards so the initial velocity must be positive..

it also makes the problem solvable, as you will soon see

max height occurs at t = -B/(2a) = -96 / (2 * -16) = -96/-32 = 3

[note that if it were -96, the max would occur at t=-3 which does not

make any sense !]

At t=3 the max height is -144 + 96(3) +16 = -144 + 288 + 16 = 160

Note again that if it was -96, this measure would also be negative..

Finally -16t^2 + 96t + 16 = 120

-16t^2 + 96t - 104 = 0

t^2 - 6t + 104/16 = 0 <--- divides by -16

t^2 - 6t + 13/2 = 0

2t^2 - 12t + 13 = 0

[12 +or- sqrt ( 144 - 4(2)(13))/ 4

[12 +or- sqrt( 144 - 104)]/4

[12 +or- sqrt( 40)]/4

[ 12 +or- 2*sqrt(10)]/ 4

[ 6 +or- sqrt(10)]/2