Solve the problem.

An object is thrown upward from the top of a 160 -foot building with an initial velocity of 48 feet per second. The height $h$ of the object after $t$ seconds is given by the quadratic equation $h=-16 t^{2}+48 t+160$. When will the object hit the ground?
$-2 \mathrm{sec}$
$5 \mathrm{sec}$
$2 \mathrm{sec}$
$160 \mathrm{sec}$

Step 1 ：The object hits the ground when the height \(h\) is zero. So, we need to solve the equation \(-16t^2 + 48t + 160 = 0\) for \(t\).

Step 2 ：This is a quadratic equation in the form \(at^2 + bt + c = 0\), where \(a = -16\), \(b = 48\), and \(c = 160\).

Step 3 ：We can solve this equation using the quadratic formula: \(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\).

Step 4 ：Substituting \(a\), \(b\), and \(c\) into the formula, we get: \(t = \frac{-48 \pm \sqrt{48^2 - 4*(-16)*160}}{2*(-16)}\)

Step 5 ：\(t = \frac{-48 \pm \sqrt{2304 + 10240}}{-32}\)

Step 6 ：\(t = \frac{-48 \pm \sqrt{12544}}{-32}\)

Step 7 ：\(t = \frac{-48 \pm 112}{-32}\)

Step 8 ：This gives us two solutions: \(t = \frac{-48 - 112}{-32} = 5\) seconds and \(t = \frac{-48 + 112}{-32} = -2\) seconds.

Step 9 ：However, time cannot be negative, so we discard the negative solution. Therefore, the object will hit the ground after \(\boxed{5}\) seconds.