Step 1 :Given the equation \(120 = -16t^2 + 156t + 19\), we rearrange it to get \(16t^2 - 156t + 101 = 0\)
Step 2 :This is a quadratic equation in the form \(at^2 + bt + c = 0\), where \(a = 16\), \(b = -156\), and \(c = 101\)
Step 3 :We can solve for \(t\) using the quadratic formula: \(t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)
Step 4 :Substituting the values gives: \(t = \frac{156 \pm \sqrt{(-156)^2 - 4*16*101}}{2*16}\)
Step 5 :Solving the equation gives: \(t = \frac{156 \pm \sqrt{17872}}{32}\)
Step 6 :This simplifies to: \(t = \frac{156 \pm 133.68}{32}\)
Step 7 :This gives two possible values for \(t\): \(t1 = \frac{156 + 133.68}{32} = 9.05\) seconds and \(t2 = \frac{156 - 133.68}{32} = 0.70\) seconds
Step 8 :\(\boxed{t = 0.70, 9.05}\)
Step 9 :To find when the object will reach the ground, we set \(h\) to 0 in the equation and solve for \(t\): \(0 = -16t^2 + 156t + 19\)
Step 10 :Rearranging the equation gives: \(16t^2 - 156t - 19 = 0\)
Step 11 :Again, we can solve for \(t\) using the quadratic formula: \(t = \frac{156 \pm \sqrt{(-156)^2 - 4*16*(-19)}}{2*16}\)
Step 12 :Solving the equation gives: \(t = \frac{156 \pm \sqrt{25552}}{32}\)
Step 13 :This simplifies to: \(t = \frac{156 \pm 159.85}{32}\)
Step 14 :This gives two possible values for \(t\): \(t1 = \frac{156 + 159.85}{32} = 9.87\) seconds and \(t2 = \frac{156 - 159.85}{32} = -0.12\) seconds
Step 15 :Since time cannot be negative, the object will reach the ground after approximately 9.87 seconds
Step 16 :\(\boxed{t = 9.87}\)