Answer :
To determine the times at which the golf ball will be on the ground, we need to find the values of [tex]\( t \)[/tex] for which [tex]\( h(t) = 0 \)[/tex]. The height function provided is:
[tex]\[ h(t) = -16(t-3)^2 + 144 \][/tex]
Step-by-step solution:
1. Set the height function equal to zero:
[tex]\[ 0 = -16(t-3)^2 + 144 \][/tex]
2. Isolate the quadratic term:
[tex]\[ 16(t-3)^2 = 144 \][/tex]
3. Divide both sides by 16:
[tex]\[ (t-3)^2 = \frac{144}{16} \][/tex]
[tex]\[ (t-3)^2 = 9 \][/tex]
4. Take the square root of both sides:
[tex]\[ t-3 = \pm3 \][/tex]
5. Solve for [tex]\( t \)[/tex] by adding 3 to both solutions:
[tex]\[ t - 3 = 3 \quad \text{or} \quad t - 3 = -3 \][/tex]
[tex]\[ t = 3 + 3 \quad \text{or} \quad t = 3 - 3 \][/tex]
[tex]\[ t = 6 \quad \text{or} \quad t = 0 \][/tex]
Thus, the golf ball will be on the ground at [tex]\( t = 0 \)[/tex] seconds and [tex]\( t = 6 \)[/tex] seconds.
The correct statement is:
- The golf ball will be on the ground at 0 and 6 seconds.
[tex]\[ h(t) = -16(t-3)^2 + 144 \][/tex]
Step-by-step solution:
1. Set the height function equal to zero:
[tex]\[ 0 = -16(t-3)^2 + 144 \][/tex]
2. Isolate the quadratic term:
[tex]\[ 16(t-3)^2 = 144 \][/tex]
3. Divide both sides by 16:
[tex]\[ (t-3)^2 = \frac{144}{16} \][/tex]
[tex]\[ (t-3)^2 = 9 \][/tex]
4. Take the square root of both sides:
[tex]\[ t-3 = \pm3 \][/tex]
5. Solve for [tex]\( t \)[/tex] by adding 3 to both solutions:
[tex]\[ t - 3 = 3 \quad \text{or} \quad t - 3 = -3 \][/tex]
[tex]\[ t = 3 + 3 \quad \text{or} \quad t = 3 - 3 \][/tex]
[tex]\[ t = 6 \quad \text{or} \quad t = 0 \][/tex]
Thus, the golf ball will be on the ground at [tex]\( t = 0 \)[/tex] seconds and [tex]\( t = 6 \)[/tex] seconds.
The correct statement is:
- The golf ball will be on the ground at 0 and 6 seconds.