Answer :
Absolutely! Let's dive into solving the problem step-by-step. We need to find out how many seconds, [tex]\( t \)[/tex], it will take for the rock to hit the water given the equation [tex]\( d = 16t^2 - 15t \)[/tex], where [tex]\( d \)[/tex] represents the distance in feet the rock falls after [tex]\( t \)[/tex] seconds. We're told that the water is 100 feet below ground, so we set [tex]\( d \)[/tex] equal to 100 and solve for [tex]\( t \)[/tex].
1. Start with the equation:
[tex]\[ d = 16t^2 - 15t \][/tex]
2. Set [tex]\( d \)[/tex] to 100:
[tex]\[ 16t^2 - 15t = 100 \][/tex]
3. Rearrange the equation to set it to zero:
[tex]\[ 16t^2 - 15t - 100 = 0 \][/tex]
4. This is now a quadratic equation of the form:
[tex]\[ at^2 + bt + c = 0 \][/tex]
where [tex]\( a = 16 \)[/tex], [tex]\( b = -15 \)[/tex], and [tex]\( c = -100 \)[/tex].
5. Solve the quadratic equation using the quadratic formula:
[tex]\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Plugging in the values of [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex]:
[tex]\[ t = \frac{-(-15) \pm \sqrt{(-15)^2 - 4 \cdot 16 \cdot (-100)}}{2 \cdot 16} \][/tex]
6. Simplify inside the square root:
[tex]\[ t = \frac{15 \pm \sqrt{225 + 6400}}{32} \][/tex]
[tex]\[ t = \frac{15 \pm \sqrt{6625}}{32} \][/tex]
7. Calculate the square root:
[tex]\[ \sqrt{6625} \approx 81.40 \][/tex]
8. Substitute back into the formula:
[tex]\[ t = \frac{15 \pm 81.40}{32} \][/tex]
9. This gives us two solutions:
[tex]\[ t = \frac{15 + 81.40}{32} \quad \text{or} \quad t = \frac{15 - 81.40}{32} \][/tex]
[tex]\[ t = \frac{96.40}{32} \quad \text{or} \quad t = \frac{-66.40}{32} \][/tex]
10. Calculate each value:
[tex]\[ t = 3.0125 \quad \text{or} \quad t = -2.075 \][/tex]
Since time [tex]\( t \)[/tex] cannot be negative, we discard [tex]\( t = -2.075 \)[/tex].
11. Round the positive solution to the nearest tenth of a second:
[tex]\[ t \approx 3.0 \][/tex]
Therefore, it will take approximately [tex]\( 3.0 \)[/tex] seconds for the rock to hit the water.
1. Start with the equation:
[tex]\[ d = 16t^2 - 15t \][/tex]
2. Set [tex]\( d \)[/tex] to 100:
[tex]\[ 16t^2 - 15t = 100 \][/tex]
3. Rearrange the equation to set it to zero:
[tex]\[ 16t^2 - 15t - 100 = 0 \][/tex]
4. This is now a quadratic equation of the form:
[tex]\[ at^2 + bt + c = 0 \][/tex]
where [tex]\( a = 16 \)[/tex], [tex]\( b = -15 \)[/tex], and [tex]\( c = -100 \)[/tex].
5. Solve the quadratic equation using the quadratic formula:
[tex]\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Plugging in the values of [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex]:
[tex]\[ t = \frac{-(-15) \pm \sqrt{(-15)^2 - 4 \cdot 16 \cdot (-100)}}{2 \cdot 16} \][/tex]
6. Simplify inside the square root:
[tex]\[ t = \frac{15 \pm \sqrt{225 + 6400}}{32} \][/tex]
[tex]\[ t = \frac{15 \pm \sqrt{6625}}{32} \][/tex]
7. Calculate the square root:
[tex]\[ \sqrt{6625} \approx 81.40 \][/tex]
8. Substitute back into the formula:
[tex]\[ t = \frac{15 \pm 81.40}{32} \][/tex]
9. This gives us two solutions:
[tex]\[ t = \frac{15 + 81.40}{32} \quad \text{or} \quad t = \frac{15 - 81.40}{32} \][/tex]
[tex]\[ t = \frac{96.40}{32} \quad \text{or} \quad t = \frac{-66.40}{32} \][/tex]
10. Calculate each value:
[tex]\[ t = 3.0125 \quad \text{or} \quad t = -2.075 \][/tex]
Since time [tex]\( t \)[/tex] cannot be negative, we discard [tex]\( t = -2.075 \)[/tex].
11. Round the positive solution to the nearest tenth of a second:
[tex]\[ t \approx 3.0 \][/tex]
Therefore, it will take approximately [tex]\( 3.0 \)[/tex] seconds for the rock to hit the water.
