Answer:
Explanation:
Given the the path of an object at a 45° with initial velocity of 80 feet per second modeled by the equation [tex]h(x) = (-\dfrac{32}{80^2} )x^2 + x[/tex] where;
x is the horizontal distance traveled and h(x) is the height in feet, if x is given as 100 ft, the height of the object will be gotten by simply substituting x = 100 into the modeled function as shown;
[tex]h(x) = (-\dfrac{32}{80^2} )x^2 + x\\\\h(100) = (-\dfrac{32}{80^2} )100^2 + 100\\\\h(100) = (-0.005 )100^2 + 100\\\\h(100) = (-0.005 )10,000 + 100\\\\h(100) = -50 + 100\\\\h(100) =50 feet[/tex]
Hence height of the object when it has traveled 100 feet away horizontally is 50 feet.