Question
A rocket is launched from a tower. The height of the rocket, y in feet, is related to the time after launch, x in seconds, by the given equation. Using this equation, find out the time at which the rocket will reach its max, to the nearest 100th of a second.
y, equals, minus, 16, x, squared, plus, 107, x, plus, 60
y=−16x
2
+107x+60
Answer
Attempt 1 out of 2
y, equals, minus, 16, x, squared, plus, 107, x, plus, 60
y=−16x
2
+107x+60
Answer
Attempt 1 out of 2
Answers
GPT-4o mini
To find the time at which the rocket reaches its maximum height, we need to determine the vertex of the quadratic function given by the equation:
\[
y = -16x^2 + 107x + 60
\]
For a quadratic equation of the form \( y = ax^2 + bx + c \), the \( x \)-coordinate of the vertex (which gives the maximum height when \( a < 0 \)) can be calculated using the formula:
\[
x = -\frac{b}{2a}
\]
In this case, the coefficients are:
- \( a = -16 \)
- \( b = 107 \)
Now plug these values into the formula:
\[
x = -\frac{107}{2 \times -16}
\]
\[
x = -\frac{107}{-32}
\]
\[
x = \frac{107}{32}
\]
Calculating \( \frac{107}{32} \):
\[
x \approx 3.34375
\]
Now, rounding to the nearest hundredth of a second:
\[
x \approx 3.34 \text{ seconds}
\]
Therefore, the time at which the rocket will reach its maximum height is approximately **3.34 seconds**.
\[
y = -16x^2 + 107x + 60
\]
For a quadratic equation of the form \( y = ax^2 + bx + c \), the \( x \)-coordinate of the vertex (which gives the maximum height when \( a < 0 \)) can be calculated using the formula:
\[
x = -\frac{b}{2a}
\]
In this case, the coefficients are:
- \( a = -16 \)
- \( b = 107 \)
Now plug these values into the formula:
\[
x = -\frac{107}{2 \times -16}
\]
\[
x = -\frac{107}{-32}
\]
\[
x = \frac{107}{32}
\]
Calculating \( \frac{107}{32} \):
\[
x \approx 3.34375
\]
Now, rounding to the nearest hundredth of a second:
\[
x \approx 3.34 \text{ seconds}
\]
Therefore, the time at which the rocket will reach its maximum height is approximately **3.34 seconds**.