Question

A rock is thrown from the top of a tall building. The distance, in feet, between the rock and the ground X seconds after it is thrown is given by f(X) = -16x^2 - 4x + 382. How long after the rock is thrown does it hit the ground?

Answer 1

Given: The function representing the distance, in feet, between the rock and the ground X seconds after it is thrown as

`f(x)=-16x^2-4x+382`

To Determine: How long after the rock is thrown does it hit the ground

Solution

Please note that when the rock hits the ground, the vertical distance f(x) would be zero

Therefore

`\begin{gathered} f(x)=0 \\ -16x^2-4x+382=0 \end{gathered}`

Solve for x in the equation using the quadratic formula

The quadratic formula is given as

`x=(-b\pm√(b^2-4ac))/(2a)`
`\begin{gathered} -16x^2-4x+382=0 \\ a=-16,b=-4,c=382 \\ x=(-(-4)\pm√((-4)^2-4*-16*382))/(2*-16) \end{gathered}`
`x=(4\pm√(16+24448))/(-32)`
`\begin{gathered} x=(4\pm√(24464))/(-32) \\ x=(4\pm156.4097183)/(-32) \end{gathered}`
`\begin{gathered} x=(4+156.4097183)/(-32),or,x=(4-156.4097183)/(-32) \\ x=(160.40971..)/(-32),or,x=(-152.40971883)/(-32) \\ x=-5.0128,or,x=4.7628 \end{gathered}`