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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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How do you find the y-coordinate vertex of a parabola?

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.


How do you find the vertexof a parabola?

To find the vertex of a parabola given its equation in standard form (y = ax^2 + bx + c), you can use the formula for the x-coordinate of the vertex: (x = -\frac{b}{2a}). Once you have the x-coordinate, substitute it back into the equation to find the corresponding y-coordinate. Thus, the vertex can be expressed as the point ((-\frac{b}{2a}, f(-\frac{b}{2a}))). For parabolas in vertex form (y = a(x-h)^2 + k), the vertex is simply the point ((h, k)).


Where is the vertex coordinate of the parabola y equals 24 -6x -3x squared when plotted on the Cartesian plane?

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.


To find the value of a in a parabola opening up or down, subtract the y-value of the parabola at the vertex from the y-value of the point on the parabola that is one unit to the of the vertex?

right


What could be the equation of a parabola with its vertex at (-36).?

The equation of a parabola with its vertex at the point (-36, k) can be expressed in the vertex form as ( y = a(x + 36)^2 + k ), where ( a ) determines the direction and width of the parabola. If the vertex is at (-36), the x-coordinate is fixed, but the y-coordinate ( k ) can vary depending on the specific position of the vertex. If you'd like a specific example, assuming ( k = 0 ) and ( a = 1 ), the equation would be ( y = (x + 36)^2 ).

Related Questions

When a parabola opens upward the y coordinate of the vertex is a what?

Opening up, the vertex is a minimum.


What is the y-coordinate of the vertex of a parabola with the following equation?

The y coordinate is given below:


To find the value of a in a parabola opening up or down subtract the y-value of the parabola at the vertex from the y-value of the point on the parabola that is one unit to the of the vertex?

To find the value of a in a parabola opening up or down subtract the y-value of the parabola at the vertex from the y-value of the point on the parabola that is one unit to the right of the vertex.


How do you find the y-coordinate vertex of a parabola?

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.


How do you find the vertexof a parabola?

To find the vertex of a parabola given its equation in standard form (y = ax^2 + bx + c), you can use the formula for the x-coordinate of the vertex: (x = -\frac{b}{2a}). Once you have the x-coordinate, substitute it back into the equation to find the corresponding y-coordinate. Thus, the vertex can be expressed as the point ((-\frac{b}{2a}, f(-\frac{b}{2a}))). For parabolas in vertex form (y = a(x-h)^2 + k), the vertex is simply the point ((h, k)).


Where is the vertex coordinate of the parabola y equals x square -3x -10 located on the Cartesian plane?

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Where is the vertex coordinate of the parabola y equals 24 -6x -3x squared when plotted on the Cartesian plane?

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.


To find the value of a in a parabola opening up or down, subtract the y-value of the parabola at the vertex from the y-value of the point on the parabola that is one unit to the of the vertex?

right


What is the y-coordinate of the vertex of the parabola that is given by the equation below?

We will be able to identify the answer if we have the equation. We can only check on the coordinates from the given vertex.


What could be the equation of a parabola with its vertex at (-36).?

The equation of a parabola with its vertex at the point (-36, k) can be expressed in the vertex form as ( y = a(x + 36)^2 + k ), where ( a ) determines the direction and width of the parabola. If the vertex is at (-36), the x-coordinate is fixed, but the y-coordinate ( k ) can vary depending on the specific position of the vertex. If you'd like a specific example, assuming ( k = 0 ) and ( a = 1 ), the equation would be ( y = (x + 36)^2 ).


Where is the vertex coordinate of the parabola y equals 24 -6x -3x squared?

To find the vertex of the parabola given by the equation (y = 24 - 6x - 3x^2), we can rewrite it in standard form (y = ax^2 + bx + c). Here, (a = -3), (b = -6), and (c = 24). The x-coordinate of the vertex can be found using the formula (x = -\frac{b}{2a}), which gives (x = -\frac{-6}{2 \cdot -3} = 1). Substituting (x = 1) back into the equation, we find the y-coordinate: (y = 24 - 6(1) - 3(1^2) = 15). Therefore, the vertex coordinate is ((1, 15)).


How do you get directrix of a parabola?

The directrix of a parabola can be found using its standard form equation. For a parabola that opens upwards or downwards, given by (y = ax^2 + bx + c), the directrix is located at (y = k - \frac{1}{4p}), where (k) is the vertex's y-coordinate and (p) is the distance from the vertex to the focus. For a parabola that opens sideways, the directrix is given by (x = h - \frac{1}{4p}), where (h) is the vertex's x-coordinate. The value of (p) can be determined based on the coefficients of the quadratic equation.