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)
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 !
+15001 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.
!
