Question

(u5) two jedi are having a training battle. one jedi (the master) runs up a wall and does a flip in the air. his jump can be modeled by the equation: $f(x) = -0.8x^2 + 4x + 6$, where $x$ equals the time in the air and $f(x)$ is the height of his jump. the other jedi (his apprentice) swings her lightsaber at her master. the swing of the other jedi’s lightsaber can be measured by $g(x) = .85x + 1.75$ where $x$ is the time in seconds and $g(x)$ is the height of her lightsaber in feet. when will the lightsaber hit the jedi?
seconds

Explanation:

Step1: Set the equations equal

To find when the lightsaber hits the Jedi, we set \( f(x) = g(x) \). So, \( -0.8x^{2}+4x + 6=0.85x + 1.75 \).

Step2: Rearrange into standard quadratic form

Subtract \( 0.85x \) and \( 1.75 \) from both sides: \( -0.8x^{2}+4x - 0.85x+6 - 1.75 = 0 \), which simplifies to \( -0.8x^{2}+3.15x + 4.25 = 0 \). Multiply through by -100 to eliminate decimals: \( 80x^{2}-315x - 425 = 0 \). Divide by 5: \( 16x^{2}-63x - 85 = 0 \).

Step3: Use quadratic formula

The quadratic formula is \( x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a} \) for \( ax^{2}+bx + c = 0 \). Here, \( a = 16 \), \( b=-63 \), \( c = -85 \). Calculate the discriminant: \( \Delta=b^{2}-4ac=(-63)^{2}-4\times16\times(-85)=3969 + 5440=9409 \). Then \( \sqrt{\Delta}=\sqrt{9409} = 97 \). So \( x=\frac{63\pm97}{2\times16} \). We have two solutions: \( x=\frac{63 + 97}{32}=\frac{160}{32}=5 \) and \( x=\frac{63-97}{32}=\frac{-34}{32}=-1.0625 \). Since time can't be negative, we take \( x = 5 \).

Answer:

5