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The height of an object is given by h = -16t^2 + 137t + 100
Find the two points in time when the object is 148 feet above the ground . Round your answer to the nearest hundredth of a second (two decimal places)

 Feb 9, 2021
 #1
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h = -16t^2 + 137t + 100

148 = -16t^2 + 137t + 100

-16t^2+137t-48 = 0               Use quadratic formula to find the two values of   t

 

a = -16     b = 137     c = -48

 

\(t = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

 

\(t = {-137 \pm \sqrt{137^2-4(-16)(-48)} \over 2(-16)}\)             caculate away ! 

 Feb 9, 2021
 #2
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The hight of an object is given by h = -16t^2 + 137t + 100
Find the two points in time when the object is 148 feet above the ground . Round your answer to the nearest hundredth of a second (two decimal places)

 

Hello Guest!

 

\(h = -16t ^ 2 + 137t + 100\\ 148= -16t ^ 2 + 137t + 100\\ -16t ^ 2 + 137t + 100-148=0\)

\( -16t ^ 2 + 137t -48=0\)

\(x = \large {-137 \pm \sqrt{137^2-4\cdot (-16)\cdot (-48)} \over 2\cdot(-16)}\\ x=\dfrac{-137\pm 125.288}{-32}\)

\(x_1=0.37\\ x_2=8.20\)

 

The object is 0.37 seconds and 8.20 seconds after being shot

from a height of 100 feet at a height of 148 feet above ground.

laugh  !

 Feb 9, 2021
edited by asinus  Feb 9, 2021
edited by asinus  Feb 9, 2021

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