Answer :
To solve the problem of finding the interval of time during which Jerald is less than 104 feet above the ground, we need to analyze the equation given for his height:
[tex]\[ h = -16t^2 + 729 \][/tex]
We are given that we want to find when [tex]\( h < 104 \)[/tex]. So, we set up the inequality:
[tex]\[ -16t^2 + 729 < 104 \][/tex]
Next, we'll simplify this inequality:
1. Subtract 729 from both sides to isolate the quadratic term:
[tex]\[ -16t^2 < 104 - 729 \][/tex]
2. Simplify the right side:
[tex]\[ -16t^2 < -625 \][/tex]
3. Divide each side by [tex]\(-16\)[/tex]. We have to be careful because dividing by a negative number means we need to reverse the inequality sign:
[tex]\[ t^2 > \frac{625}{16} \][/tex]
4. Take the square root of both sides to solve for [tex]\( t \)[/tex]. Since we’re dealing with time, we only consider the positive time values:
[tex]\[ t > \sqrt{\frac{625}{16}} \][/tex]
[tex]\[ t > \frac{25}{4} \][/tex]
[tex]\[ t > 6.25 \][/tex]
Since negative time values don't make sense in this context, we only consider the times greater than [tex]\( 6.25 \)[/tex]. Therefore, Jerald is less than 104 feet above the ground after 6.25 seconds.
The correct interval of time is:
[tex]\[ t > 6.25 \][/tex]
Therefore, the appropriate answer from the given options is:
[tex]\[ t > 6.25 \][/tex]
[tex]\[ h = -16t^2 + 729 \][/tex]
We are given that we want to find when [tex]\( h < 104 \)[/tex]. So, we set up the inequality:
[tex]\[ -16t^2 + 729 < 104 \][/tex]
Next, we'll simplify this inequality:
1. Subtract 729 from both sides to isolate the quadratic term:
[tex]\[ -16t^2 < 104 - 729 \][/tex]
2. Simplify the right side:
[tex]\[ -16t^2 < -625 \][/tex]
3. Divide each side by [tex]\(-16\)[/tex]. We have to be careful because dividing by a negative number means we need to reverse the inequality sign:
[tex]\[ t^2 > \frac{625}{16} \][/tex]
4. Take the square root of both sides to solve for [tex]\( t \)[/tex]. Since we’re dealing with time, we only consider the positive time values:
[tex]\[ t > \sqrt{\frac{625}{16}} \][/tex]
[tex]\[ t > \frac{25}{4} \][/tex]
[tex]\[ t > 6.25 \][/tex]
Since negative time values don't make sense in this context, we only consider the times greater than [tex]\( 6.25 \)[/tex]. Therefore, Jerald is less than 104 feet above the ground after 6.25 seconds.
The correct interval of time is:
[tex]\[ t > 6.25 \][/tex]
Therefore, the appropriate answer from the given options is:
[tex]\[ t > 6.25 \][/tex]
