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it
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11. Timmy is playing in a
dangerous old well. He
throws a rock down the well.
The distance d (in feet) the
rock falls after t seconds can
be represented by the
equation d=16″ – 15r. If
the water is 100 feet below
ground, how many seconds
will it take for the rock to
hit the water? Round your
answer the the nearest tenth
of a second.



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.