Answer:
Part A)
Approximately 318.1318 meters.
Part B)
Approximately 137.7551 meters.
Step-by-step explanation:
The path of a projectile is given by the equation:
[tex]\displaystyle y = \sqrt{3} x -\frac{49x^2}{9000}[/tex]
Part A)
The range of the projectile will be given by the difference between its starting point and landing point. In other words, its two zeros.
Let y = 0 and solve for x:
[tex]\displaystyle 0 = \sqrt{3}x - \frac{49x^2}{9000}[/tex]
Factor:
[tex]\displaystyle 0 = x\left(\sqrt{3} - \frac{49x}{9000}\right)[/tex]
Zero Product Property:
[tex]\displaystyle x = 0 \text{ or } \sqrt{3} - \frac{49x}{9000} = 0[/tex]
Solve for each case:
[tex]\displaystyle x = 0 \text{ or } x = \frac{9000\sqrt{3}}{49}\approx 318.1318[/tex]
Hence, the range of the projectile is approximately (318.1318 - 0) or 318.1318 meters.
Part B)
Since the equation is a quadratic, the maximum height is given by its vertex. Recall that the vertex of a quadratic is given by:
[tex]\displaystyle \text{Vertex} = \left(-\frac{b}{2a}, f\left(-\frac{b}{2a}\right)\right)[/tex]
In this case, a = -49/9000 and b = √3.
Find the x-coordinate of the vertex:
[tex]\displaystyle x = - \frac{(\sqrt{3})}{2\left(\dfrac{-49}{9000}\right)} = \frac{4500\sqrt{3}}{49}[/tex]
Then the maximum height will be:
[tex]\displaystyle \displaystyle \begin{aligned} y\left(\frac{4500\sqrt{3}}{49}\right) &=\sqrt{3} \left(\frac{4500\sqrt{3}}{49}\right) -\frac{49\left(\dfrac{4500\sqrt{3}}{49}\right)^2}{9000} \\ \\ &= \frac{13500}{49} -\frac{6750}{49}\\ \\ &=\frac{6750}{49}\\ \\ &\approx 137.7551\text{ meters}\end{aligned}[/tex]
The maximum height reached by the projectile will be 137.7551 meters.
