An object is thrown upward at a speed of 156 feet per second by a machine from a height of 19 feet off the ground. The height $h$ of the object after $t$ seconds can be found using the equation
\[
h=-16 t^{2}+156 t+19
\]
(Tip: approximate all numbers to two decimal places.)

When will the height be 120 feet?

Hint: Set $h$ to 120 .

When will the object reach the ground?
Select an answer

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 ：\(\boxed{t = 9.87}\)