To answer the given questions, we need to solve the equation for specific conditions. The height [tex]\( h \)[/tex] of an object thrown upward is given by:
[tex]\[ h = -16t^2 + 146t + 9 \][/tex]
### 1. When will the height be 243 feet?
We need to find the time [tex]\( t \)[/tex] when the height [tex]\( h \)[/tex] is 243 feet. Plugging 243 into the height equation, we get:
[tex]\[ 243 = -16t^2 + 146t + 9 \][/tex]
Let's rearrange the equation to standard quadratic form:
[tex]\[ -16t^2 + 146t + 9 - 243 = 0 \][/tex]
[tex]\[ -16t^2 + 146t - 234 = 0 \][/tex]
This quadratic equation can be solved for [tex]\( t \)[/tex] using the quadratic formula [tex]\( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = -16 \)[/tex], [tex]\( b = 146 \)[/tex], and [tex]\( c = -234 \)[/tex].
After solving the quadratic equation, the solutions for [tex]\( t \)[/tex] are:
[tex]\[ t = \frac{73}{16} - \frac{\sqrt{1585}}{16} \quad \text{and} \quad t = \frac{\sqrt{1585}}{16} + \frac{73}{16} \][/tex]
### 2. When will the object reach the ground?
The object reaches the ground when its height [tex]\( h \)[/tex] is 0 feet. So we need to solve the equation:
[tex]\[ 0 = -16t^2 + 146t + 9 \][/tex]
Since this is already in the standard quadratic form, we can use the quadratic formula again with [tex]\( a = -16 \)[/tex], [tex]\( b = 146 \)[/tex], and [tex]\( c = 9 \)[/tex].
After solving this quadratic equation, the solutions for [tex]\( t \)[/tex] are:
[tex]\[ t = \frac{73}{16} - \frac{\sqrt{5473}}{16} \quad \text{and} \quad t = \frac{\sqrt{5473}}{16} + \frac{73}{16} \][/tex]
Thus, these are the times when the object reaches the given heights.
