Example:
Vertical velocity component of a 100 m/s cannonball fired at 30 degrees from horizontal = sin 30 * 100 = 50 m/s
Find the maximum height reached.
Take acceleration due to gravity at 10 (m/s)/s
so:
u = 50 m/s
v = 0
a = - 10 m/s/s (approx)
s = ?
Using:
v2 = u2 + ( 2 * a * s )
so:
s = ( v2 - u2 ) / ( 2 * a )
s = ( 0 - 2500 ) / ( 2 * -10 )
s = 125 metres
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Once you calculate the X coordinate using the axis of symmetry (X=-b/2a), you plug that value in for all of the X's in the equation of the parabola. You then solve the equation for the value of Y.
The vertex coordinate point of the vertex of the parabola y = 24-6x-3x^2 when plotted on the Cartesian plane is at (-1, 27) which can also be found by completing the square.
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In a quadratic y = ax² + bx + c, the roots are where y = 0, and the parabola crosses the x-axis. The average of these two roots is the x coordinate of the vertex of the parabola.
Suppose the equation of the parabola is y = ax2 + bx + c where a, b, and c are constants, and a ≠0. The roots of the parabola are given by x = [-b ± sqrt(D)]/2a where D is the discriminant. Rather than solve explicitly for the coordinates of the vertex, note that the vertical line through the vertex is an axis of symmetry for the parabola. The two roots are symmetrical about x = -b/2a so, whatever the value of D and whether or not the parabola has real roots, the x coordinate of the vertex is -b/2a. It is simplest to substitute this value for x in the equation of the parabola to find the y-coordinate of the vertex, which is c - b2/2a.