Asked by Jalen
A snowball was thrown from a cliff. The height above ground of the snowball is modeled by the function h(t) = -16t2 + 48t + 28, where h is height in feet, and t is time in seconds. How long was the snowball in the air?
Did everything got (4t+7)(t+1)
then did 4t+7=0 i got t=-1.75 then i did the x+1=0 i got -1, its asking for how long it was in the air(seconds) what did i do wrong or how do i find the answer.
Did everything got (4t+7)(t+1)
then did 4t+7=0 i got t=-1.75 then i did the x+1=0 i got -1, its asking for how long it was in the air(seconds) what did i do wrong or how do i find the answer.
Answers
Answered by
Reiny
Snowball was in the air as long as h > 0
let's see when h = 0
0 = -16t^2 + 48t + 28
divide by -4
4t^2 - 12t - 7 = 0
(2t + 1)(2t - 7) = 0
so t = -1/2 or t = 7/2
but t ≥ 0
so the ball was in the air for 7/2 or 3.5 seconds
let's see when h = 0
0 = -16t^2 + 48t + 28
divide by -4
4t^2 - 12t - 7 = 0
(2t + 1)(2t - 7) = 0
so t = -1/2 or t = 7/2
but t ≥ 0
so the ball was in the air for 7/2 or 3.5 seconds
Answered by
Shelby
If you plug it in to a calculator you get 3.5
You didn't factor correctly
(4t+7)(t+1)=4t^2+11t+7
-16t^2+48t+28
-16t^2-8t+56t+28
(16t+8)(-t+3.5)
t=-1/2 t=3.5
You didn't factor correctly
(4t+7)(t+1)=4t^2+11t+7
-16t^2+48t+28
-16t^2-8t+56t+28
(16t+8)(-t+3.5)
t=-1/2 t=3.5
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