An object is fired vertically upward with an initial speed of 68.6 meters/second. After t seconds, the height is shown by
y = -49 t^2 + 68.6 t meters. Given that this object has an initial speed of 68.6 m/s, what is the maximum height it will reach?
oops, I made a mistake.
y= -4.9t^2 + 68.6t
See
http://www.jiskha.com/display.cgi?id=1369047938
To find the maximum height that the object will reach, we need to determine the vertex of the parabolic function given by the equation y = -49t^2 + 68.6t.
The vertex of a parabola is given by the formula (h, k), where h represents the x-coordinate and k represents the y-coordinate.
To find the x-coordinate of the vertex, we can use the formula h = -b / (2a), where a and b are the coefficients of the quadratic equation in the form y = at^2 + bt + c. In this case, a = -49 and b = 68.6.
h = -b / (2a) = -68.6 / (2 * -49) = 0.7
So the x-coordinate of the vertex is h = 0.7.
To find the y-coordinate of the vertex, we substitute the value of h into the equation y = -49t^2 + 68.6t:
k = -49(0.7)^2 + 68.6(0.7) = -24.465 + 48.02 = 23.555
Therefore, the maximum height that the object will reach is 23.555 meters.