An object is thrown upward at a speed of 190 feet per second by a machine from a height of 6 feet off the ground. The height [tex]h[/tex] of the object after [tex]t[/tex] seconds can be found using the equation:

[tex]\[ h = -16t^2 + 190t + 6 \][/tex]

1. When will the height be 336 feet?
[tex]\[ \square \text{ Select an answer } \checkmark \][/tex]

2. When will the object reach the ground?
[tex]\[ \square \text{ Select an answer } \checkmark \][/tex]

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Answer :

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]