A ball is thrown upward and outward from a height of 5 feet. The height of the ball, $$ f(x) $$, in feet, can be modeled by $$ f(x)=-0.1 x^{2}+0.6 x+5 $$ where $$ x $$ is the ball's horizontal distance, in feet, from where it was thrown. Use this model to solve parts (a) through (c). a. What is the maximum height of the ball and how far from where it was thrown does this occur? The maximum height is 5.9 feet, which occurs 3.0 feet from the point of release. (Round to the nearest tenth as needed.) b. How far does the ball travel horizontally before hitting the ground? $$ \square $$ feet (Round to the nearest tenth as needed.)

Question

UpStudy AI · Accepted Answer

We start with the height function  
$$
f(x) = -0.1x^2 + 0.6x + 5
$$
To find how far the ball travels horizontally before hitting the ground, we need to determine the value of $$ x $$ when $$ f(x) = 0 $$.

1. Set up the equation:
   $$
   -0.1x^2 + 0.6x + 5 = 0
   $$

2. Multiply each term by $$-10$$ to eliminate decimals:
   $$
   x^2 - 6x - 50 = 0
   $$

3. Use the quadratic formula:
   $$
   x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
   $$
   In our equation, $$ a = 1 $$, $$ b = -6 $$, and $$ c = -50 $$.

4. Substitute the values into the formula:
   $$
   x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(1)(-50)}}{2(1)} = \frac{6 \pm \sqrt{36 + 200}}{2} = \frac{6 \pm \sqrt{236}}{2}
   $$

5. Simplify the expression:
   $$
   x = \frac{6 \pm \sqrt{236}}{2}
   $$
   Since $$\sqrt{236} \approx 15.4$$, we have:
   $$
   x = \frac{6 \pm 15.4}{2}
   $$

6. Since distance cannot be negative, we take the positive root:
   $$
   x = \frac{6 + 15.4}{2} \approx \frac{21.4}{2} \approx 10.7
   $$

Thus, the ball travels approximately $$10.7$$ feet horizontally before hitting the ground.
The ball travels approximately 10.7 feet horizontally before hitting the ground.

Solución de inteligencia artificial de Upstudy

Responder

Solución

Extra Insights

preguntas relacionadas

Latest Pre Calculus Questions