An emergency flare is shot vertically into the air with a speed of 60m/s. It's height h metres after t seconds is given by the equation h=−5(t−6)^2+180
To the nearest second, when does the flare reach its maximum height?
impatient much?
see your previous post.
I know but this one is To the nearest second the other one was just the maximum height. Would the equation still be the same or would it be a different equation?
The vertex of the parabola is the top when t = 6 and h = 180
To find the time at which the flare reaches its maximum height, we need to determine the vertex of the quadratic equation that represents the height of the flare. The vertex form of a quadratic equation is given by:
h = a(t - h)^2 + k
In this case, the equation for the height of the flare is h = -5(t - 6)^2 + 180. Comparing this equation to the vertex form, we can see that a = -5, h = 6, and k = 180.
The time at which the flare reaches its maximum height corresponds to the x-coordinate of the vertex. The x-coordinate of the vertex is given by the formula:
t = -b / (2a)
In this case, b = h = 6 and a = -5. Plugging these values into the formula, we get:
t = -6 / (2 * -5)
t = -6 / -10
t = 0.6
Therefore, the flare reaches its maximum height approximately 0.6 seconds after it's shot into the air. When rounded to the nearest second, the flare reaches its maximum height after 1 second.