Answered

Noelle stands at the edge of a cliff and drops a rock. The height of the rock, in meters, is given by the function:

[tex]\[ f(x) = -4.9x^2 + 17 \][/tex]

where [tex]\( x \)[/tex] is the number of seconds after Noelle releases her rock.

Cesar, standing nearby on the ground, throws a rock straight up. The height of Cesar's rock, in meters, is given by the function:

[tex]\[ g(x) = -4.9x^2 + 13x \][/tex]

where [tex]\( x \)[/tex] is the number of seconds after he releases his rock.

There is a moment when the rocks are at the same height.

What is this height?
Enter your answer, rounded to the nearest tenth of a meter.

[tex]\[ \square \, \text{m} \][/tex]



Answer :

GinnyAnswer
To find the height at which Noelle's and Cesar's rocks are at the same height, we need to determine the time [tex]\( x \)[/tex] when their heights are equal, and then evaluate the height at this time.

The functions describing the heights of the rocks are:
[tex]\[ f(x) = -4.9x^2 + 17 \][/tex]
[tex]\[ g(x) = -4.9x^2 + 13x \][/tex]

We set these functions equal to each other to find the time [tex]\( x \)[/tex] when both rocks are at the same height:
[tex]\[ -4.9x^2 + 17 = -4.9x^2 + 13x \][/tex]

Subtracting [tex]\(-4.9x^2\)[/tex] from both sides, we get:
[tex]\[ 17 = 13x \][/tex]

Solving for [tex]\( x \)[/tex], we divide both sides by 13:
[tex]\[ x = \frac{17}{13} \][/tex]

Next, we substitute [tex]\( x = \frac{17}{13} \)[/tex] back into either height function to find the height. Using the function [tex]\( f(x) \)[/tex]:
[tex]\[ f\left(\frac{17}{13}\right) = -4.9\left(\frac{17}{13}\right)^2 + 17 \][/tex]

We calculate this value:
[tex]\[ \left(\frac{17}{13}\right)^2 = \frac{289}{169} \][/tex]
[tex]\[ -4.9 \times \frac{289}{169} = -4.9 \times 1.7100591716 \approx -8.3792899408284 \][/tex]
[tex]\[ f \left(\frac{17}{13}\right) = -8.3792899408284 + 17 \approx 8.6207100591716 \][/tex]

Rounding this to the nearest tenth of a meter, we get:
[tex]\[ \text{Height} \approx 8.6 \text{ meters} \][/tex]

Thus, the height at which both rocks are the same is approximately:
[tex]\[ 8.6 \text{ meters} \][/tex]

Other Questions