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Hello I need help with a question if you are able to assist.

Hello I need help with a question if you are able to assist.-example-1
User Aaron
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1 Answer

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We are given an equation that represents the path of the longest shot put by the women's track team as:


h(x)=-0.017x^2\text{ + 1.08x + 5.8}

Where x represents the horizontal distance from the start and h(x) is the height of the shot put above the ground.

Solution:

(a) The vertex of the model:

Method 1: We can plot the model on a graph and identify the vertex from the graph. The vertex is the lowest/highest point on the curve.

The graph is shown below:

From the graph, we can see that the vertex of the model is : (31.765, 22.953)

Method 2: The vertex of the model can also be calculated using the formula:

First, we find the x-coordinate using the formula:


x\text{ = -}(b)/(2a)

Substituting, we have:


\begin{gathered} x\text{ = -}\frac{1.08}{2\text{ }*\text{ -0.017}} \\ x\text{ = 31.765} \end{gathered}

The y-coordinate of the vertex can be obtained by substituting the x-coordinate into the model. So, we have the y-coordinate to be:


\begin{gathered} y\text{ = h(31.765)} \\ y=-0.017(31.765)^2\text{ + 1.08(31.765) + 5.8} \\ y\text{ = 22.953} \end{gathered}

Hence, the vertex is : (31.765, 22.953)

(b) The maximum height

The horizontal distance of the maximum height of the shot put from the starting point is :

From the plot, the starting point is -4.98


\begin{gathered} \text{Horizontal distance = 31.765 + 4.98} \\ =\text{ 36.745 f}eet \end{gathered}

The maximum height of the shot put is 22.953 feet, 36.745 feet away from the starting point.

(c) The vertical intercept

The vertical intercept is the point where the curve cuts the y-axis.

From the graph, the point is (0, 5.8).

At this point where the horizontal distance is 4.98 feet away from the starting point, the shot put is 5.8 feet above the ground.

(d) The distance from the starting point to when the shot-put stikes the ground is:

From the plot, the intercepts on the x-axis are (-4.98, 0) and (68.509, 0).

Hence, the distance is:


\begin{gathered} =\text{ 68.509 + 4.98} \\ =\text{ 73.489} \\ \cong\text{ 73.49 f}eet\text{ (2.d.p)} \end{gathered}

Hello I need help with a question if you are able to assist.-example-1
User Thlim
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