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I need help finding how far the ball travels horizontally before hitting the ground

I need help finding how far the ball travels horizontally before hitting the ground-example-1
User Xavlours
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1 Answer

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Answer

The horizontal distance (x) when the ball hits the ground is 12.8 feet

SOLUTION

Problem Statement

The question tells us that a ball thrown from a height of 6 feet, is modeled by the following equation:


\begin{gathered} f(x)=-0.2x^2+2.1x+6 \\ \text{where,} \\ f(x)\text{ is the height of the ball} \\ x\text{ is the horizontal distance covered by the ball} \end{gathered}

We are asked to find the horizontal distance covered by the ball before hitting the ground.

Method

- To find the horizontal distance covered before hitting the ground implies that the height of the ball after some horizontal distance covered must be zero.

- This is because for the ball to hit the ground, the height of the ball over the earth must be zero. This means that:


f(x)=0

- Thus, to find the horizontal distance x when the ball hits the ground, then we need to equate the function of f(x) to zero and then solve for x.

That is:


\begin{gathered} \text{Solve,} \\ -0.2x^2+2.1x+6=0 \end{gathered}

Implementation


\begin{gathered} -0.2x^2+2.1x+6=0 \\ \text{ We can solve this using the Quadratic Formula:} \\ \\ \text{The Quadratic formula is defined below:} \\ \text{Given the following Quadratic Equation:} \\ ax^2+bx+c=0 \\ \\ \text{The Quadratic formula is:} \\ x=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ \\ \text{From the equation given, we can see that:} \\ a=-0.2,b=2.1,c=6 \\ \\ \therefore x=\frac{-2.1\pm\sqrt[]{2.1^2-4(-0.2)(6)}}{2(-0.2)} \\ \\ x=(-2.1\pm3.0348)/(-0.4) \\ \\ x=(-2.1-3.0348)/(-0.4)\text{ or }(-2.1+3.0348)/(-0.4) \\ \\ \\ \therefore x=12.837\text{ or }-2.337. \\ \\ \text{ Since a horizontal distance cannot be negative, we have that:} \\ x=12.837\approx12.8\text{ (to the nearest tenth)} \end{gathered}

Final Answer

The horizontal distance (x) when the ball hits the ground is 12.8 feet

User Hasrthur
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