Questions: 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 the maximum height reached by the rocket, to the nearest tenth of a foot.
y=-16 x^2+198 x+77
Transcript text: 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 the maximum height reached by the rocket, to the nearest tenth of a foot.
\[
y=-16 x^{2}+198 x+77
\]
Solution
Solution Steps
Step 1: Identify the Type of Equation
The given equation is a quadratic equation of the form:
y=−16x2+198x+77
This is a standard form of a quadratic equation, ax2+bx+c, where a=−16, b=198, and c=77.
Step 2: Determine the Vertex of the Parabola
The maximum height of the rocket corresponds to the vertex of the parabola, since the parabola opens downwards (as indicated by the negative coefficient of x2).
The x-coordinate of the vertex can be found using the formula:
x=−2ab
Substituting the values of a and b:
x=−2×−16198=32198=6.1875
Step 3: Calculate the Maximum Height
Substitute x=6.1875 back into the original equation to find the maximum height y:
y=−16(6.1875)2+198(6.1875)+77
Calculate (6.1875)2:
(6.1875)2=38.2852
Substitute back into the equation:
y=−16×38.2852+198×6.1875+77
Calculate each term:
y=−612.5632+1225.3125+77
Add the terms:
y=689.7493
Final Answer
The maximum height reached by the rocket is approximately 689.7 feet.