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A long jumper leaves the ground at an angle of [tex]$22^{\circ}$[/tex] above the horizontal, at a speed of [tex]$10 \, \text{m/s}$[/tex]. The height of the jumper can be modeled by [tex]$h(x) = -0.05 x^2 + 0.363 x$[/tex], where [tex]$h$[/tex] is the jumper's height in meters and [tex]$x$[/tex] is the horizontal distance from the point of launch.

(a) At what horizontal distance from the point of launch does the maximum height occur? Round to 2 decimal places.

The long jumper reaches a maximum height when the horizontal distance from the point of launch is approximately [tex]$\square$[/tex] meters.


Sagot :

To determine the horizontal distance at which the long jumper reaches their maximum height, we need to analyze the quadratic equation modeling the height of the jumper:

[tex]\[ h(x) = -0.05x^2 + 0.363x. \][/tex]

This equation is in the standard form of a quadratic equation:

[tex]\[ h(x) = ax^2 + bx + c, \][/tex]

where \(a = -0.05\), \(b = 0.363\), and \(c\) is implicitly zero in this context (although it is not relevant for finding the vertex).

The vertex of a parabola represented by the quadratic equation \( ax^2 + bx + c \) gives us the maximum or minimum point of the parabola. For parabolas that open downwards, like this one (since \(a < 0\)), the vertex represents the maximum point.

The x-coordinate of the vertex can be found using the formula:

[tex]\[ x = -\frac{b}{2a}. \][/tex]

Substituting the values of \(a\) and \(b\):

[tex]\[ a = -0.05, \][/tex]
[tex]\[ b = 0.363, \][/tex]

we get:

[tex]\[ x = -\frac{0.363}{2 \times -0.05}. \][/tex]

Calculating the value:

[tex]\[ x = -\frac{0.363}{-0.1} = 3.63. \][/tex]

Therefore, the horizontal distance at which the maximum height occurs is approximately:

[tex]\[ x \approx 3.63 \text{ meters}. \][/tex]

So, the long jumper reaches a maximum height when the horizontal distance from the point of launch is approximately [tex]\(\boxed{3.63}\)[/tex] meters.
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