a rocket is launched from a tower. the height of the rocket, y in feet, is related to the time after launch…

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.\n\n$y = -16x^{2}+210x + 75$

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.\n\n$y = -16x^{2}+210x + 75$

Answer

Explanation:

Step1: Identify the coefficients

The quadratic - function is (y = ax^{2}+bx + c), where (a=-16), (b = 210), (c = 75).

Step2: Find the x - coordinate of the vertex

The x - coordinate of the vertex of a quadratic function (y = ax^{2}+bx + c) is given by (x=-\frac{b}{2a}). Substitute (a=-16) and (b = 210) into the formula: [x=-\frac{210}{2\times(-16)}=\frac{210}{32}=\frac{105}{16}=6.5625]

Step3: Find the y - coordinate of the vertex

Substitute (x = 6.5625) into the equation (y=-16x^{2}+210x + 75). [y=-16\times(6.5625)^{2}+210\times6.5625 + 75] [y=-16\times43.0078125+1378.125 + 75] [y=-688.125+1378.125 + 75] [y=765]

Answer:

765.0