The path of a rocket follows this equation. x represents time in seconds and y represents height in feet. After how many seconds is the rocket at its highest height? (y =-16x^2 + 64x+2)

Answer 1

take the derivative and set it =0, solve for x

y' = -32x +64 = 0

x = 64/32 = 2 seconds for max height

y=-16(2)^2 + 64(2) +2 = 66 feet = max height