Question
while hiking, Julia kicked a small rock off a ledge that was meters above a crevasse. use the Texas Demos calculator to graph the situation using the formula y=-4.9t^2+9. use the graph to determine when the rock will hit the bottom of the crevasse, 2 meters below ground level. Round your answer to the nearest tenth of a second.
The rocket will hit the bottom of the crevasse in ______ seconds.
The rocket will hit the bottom of the crevasse in ______ seconds.
Answers
Answered by
GPT-4o mini
To determine when the rock will hit the bottom of the crevasse using the formula \( y = -4.9t^2 + 9 \), we first need to set \( y \) to the height of the bottom of the crevasse, which is 2 meters below ground level. Since the initial height is 9 meters above the ground, the equation becomes:
\[
-4.9t^2 + 9 = -2
\]
Now, rearranging the equation gives:
\[
-4.9t^2 + 9 + 2 = 0
\]
\[
-4.9t^2 + 11 = 0
\]
Next, we isolate \( t^2 \):
\[
-4.9t^2 = -11
\]
\[
t^2 = \frac{11}{4.9}
\]
Now we can calculate \( t^2 \):
\[
t^2 \approx 2.2449
\]
Taking the square root of both sides to find \( t \):
\[
t \approx \sqrt{2.2449} \approx 1.5
\]
Therefore, the rock will hit the bottom of the crevasse in approximately 1.5 seconds.
Rounding to the nearest tenth of a second, the answer is:
**The rock will hit the bottom of the crevasse in 1.5 seconds.**
\[
-4.9t^2 + 9 = -2
\]
Now, rearranging the equation gives:
\[
-4.9t^2 + 9 + 2 = 0
\]
\[
-4.9t^2 + 11 = 0
\]
Next, we isolate \( t^2 \):
\[
-4.9t^2 = -11
\]
\[
t^2 = \frac{11}{4.9}
\]
Now we can calculate \( t^2 \):
\[
t^2 \approx 2.2449
\]
Taking the square root of both sides to find \( t \):
\[
t \approx \sqrt{2.2449} \approx 1.5
\]
Therefore, the rock will hit the bottom of the crevasse in approximately 1.5 seconds.
Rounding to the nearest tenth of a second, the answer is:
**The rock will hit the bottom of the crevasse in 1.5 seconds.**
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