To solve these problems, let's first thoroughly understand the given information.
### Problem 1: When will the height be 336 feet?
We are given the height equation of the object:
[tex]\[ h = -16t^2 + 190t + 6 \][/tex]
To find the time [tex]\( t \)[/tex] when the object reaches a height of 336 feet:
[tex]\[ 336 = -16t^2 + 190t + 6 \][/tex]
We need to solve the quadratic equation:
[tex]\[ -16t^2 + 190t + 6 = 336 \][/tex]
Rearrange the equation to standard form:
[tex]\[ -16t^2 + 190t + 6 - 336 = 0 \][/tex]
[tex]\[ -16t^2 + 190t - 330 = 0 \][/tex]
This quadratic equation can be solved using the quadratic formula [tex]\( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = -16 \)[/tex], [tex]\( b = 190 \)[/tex], and [tex]\( c = -330 \)[/tex].
Solving this, we find two solutions:
[tex]\[ t \approx 2.1127 \][/tex]
[tex]\[ t \approx 9.7623 \][/tex]
So, the object will be at a height of 336 feet at approximately [tex]\( t = 2.1127 \)[/tex] seconds and [tex]\( t = 9.7623 \)[/tex] seconds.
### Problem 2: When will the object reach the ground?
To find when the object reaches the ground, set the height [tex]\( h \)[/tex] to 0:
[tex]\[ 0 = -16t^2 + 190t + 6 \][/tex]
Now we have the quadratic equation:
[tex]\[ -16t^2 + 190t + 6 = 0 \][/tex]
Solving this using the quadratic formula [tex]\( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = -16 \)[/tex], [tex]\( b = 190 \)[/tex], and [tex]\( c = 6 \)[/tex]:
We find two solutions:
[tex]\[ t \approx -0.0315 \][/tex]
[tex]\[ t \approx 11.9065 \][/tex]
Since time [tex]\( t \)[/tex] cannot be negative in this physical context, we discard the negative solution. Therefore, the object reaches the ground at approximately [tex]\( t = 11.9065 \)[/tex] seconds.
### Summary of Solutions:
1. When will the height be 336 feet?
[tex]\[ \boxed{2.1127 \text{ seconds and } 9.7623 \text{ seconds}} \][/tex]
2. When will the object reach the ground?
[tex]\[ \boxed{11.9065 \text{ seconds}} \][/tex]
