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.