U3.L1.HW: Functions & Function Notation: Question 21

Question

An object is thrown upward at a speed of 109 feet per second by a machine from a height of 20 feet off the ground. The height h of the object after t seconds can be found using the equation h = -16t² + 109t + 20  

When will the height be 144 feet?

When will the object reach the ground?

Answer 1

h(t) = -16t^2 +109t +20 = 144

-16t^2 +109t - 124 = 0

t = 109/32 + sqr(109^2 -64(124))/32

= (109 +/-sqr(109^2 -7936)/32

=(109+/-sqr3945)/32

= 109+/-57.836)/32

= 166.836/32 or 51.164/32

=about 5.2 or 1.6 seconds

to reach 144 feet as he falls in a downward and upward direction

for 3.6 seconds it's above 144 ft.

109/32 = about 3.4, approximately midpoint between when it hit 144 feet

double it add a little bit, to get when it hits the ground, maybe close to 7 seconds

max height = about -16(3.3)^2 +109(3.3) +20 = -174.24 + 359.7 +20= 185.46 +20 = 205.46 feet+ a little

it hits the ground when h(t) = 0

-16t^2 + 109t +20=0

t = 109/32 + (1/32)sqr(109^2 +1280)

= 109/32 + (1/32)sqr(11881+1280)

= (109 + sqr(

t = (109 + sqr13161)/32

t= about (109+114.72)/32

t = about 223.72/32

= about 6.99 seconds to hit the ground