An object launched from the ground has a height(in meters) after t seconds that is modeled by the equation y=-4.9t^2+58.8t
What is the maximum height of the object?
the vertex of a quadratic is at t = -b/2a = 58.8/9.8
now evaluate y at that value of t
To find the maximum height of the object, we need to determine the vertex of the quadratic equation. The vertex form of a quadratic equation is given by y = a(t - h)² + k, where (h, k) represents the coordinates of the vertex.
In the given equation, y = -4.9t^2 + 58.8t, we can see that a = -4.9.
To find the vertex, we can use the formula:
h = -b/2a
Here, b = 58.8 and a = -4.9.
h = -58.8 / (2 * -4.9)
h = -58.8 / -9.8
h = 6
So, the x-coordinate of the vertex is h = 6.
To find the y-coordinate of the vertex, we substitute the value of h into the equation:
y = -4.9(6)^2 + 58.8(6)
y = -4.9(36) + 352.8
y = -176.4 + 352.8
y = 176.4
Therefore, the maximum height of the object is 176.4 meters.
To find the maximum height of the object, we need to determine the vertex of the quadratic equation.
The general form of a quadratic equation is y = ax^2 + bx + c, where:
a = -4.9 (coefficient of the squared term)
b = 58.8 (coefficient of the linear term)
c = 0 (constant term, since there is no y-intercept)
The x-coordinate of the vertex of a quadratic equation of the form y = ax^2 + bx + c is given by:
x = -b / (2a)
In this case, x = -58.8 / (2 * -4.9) = 6
To find the y-coordinate of the vertex, substitute the value of x back into the equation:
y = -4.9(6)^2 + 58.8(6)
y = -4.9(36) + 352.8
y = -176.4 + 352.8
y = 176.4
Therefore, the maximum height of the object is 176.4 meters.