On some other planet, Thor stands on a bridge and throws his hammer at his brother Loki, who is at ground level. The hammer’s height h (in meters above the ground), t seconds after Thor threw it, is modeled by:





Answer:
h(0) = 3600
t=60 when it hits the ground
max height at 4050
at 15 seconds
Step-by-step explanation:
h(t) = -2t^2 + 60t +3600
When it is thrown is t=0
h(0) = -2(0) +60(0) +3600 = 3600
When it hits the ground is h(t) =0
0 = Â -2t^2 + 60t +3600
Factor out -2
0 = -2 (t^2 -30t -1800)
Divide by -2
0 = (t^2 -30t -1800)
Factor. Â What 2 numbers multiply to -1800 and add to -30
0 = (t+30) (t-60)
Using the zero product property
t+30 = 0 Â t-60 =0
t = -30 Â Â t = 60
Since we do not have negative time
t=60
The maximum occurs halfway between the zeros
(-30+60)/2 = 30/2 = 15
Put this into the function to find the maximum value
h(15) = -2(15)^2 + 60(15) +3600
    = -2 (225) +900+3600
    =4050
Answer:
Thrown at: 3600
Ground after: 60
Max height: 4050
After: 15
Step-by-step explanation:
h(t) = -2t² + 60t + 3600
h(0) = 3600
-2t² + 60t + 3600 = 0
t² - 30t - 1800 = 0
t² - 60t + 30t - 1800 = 0
t(t - 60) + 30(t - 60) = 0
(t - 60)(t + 30) = 0
t = 60, -30(not possible)
h(t) = -2t² + 60t + 3600
-2(t² - 30t) + 3600
-2(t² - 2(t)(15) + 15² - 15²) + 3600
-2[(t - 15)² - 225] + 3600
-2(t - 15)² + 450 + 3600
-2(t - 15)² + 4050