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Alain throws a stone off a bridge into a river below. The stone's height (in meters above the water), x seconds after Alain threw it, is modeled by: h(x)=-5x^(2)+10 x+15 How many seconds after being thrown will the stone hit the water? seconds

Alain throws a stone off a bridge into a river below.\newlineThe stone's height (in meters above the water), xx seconds after Alain threw it, is modeled by:\newlineh(x)=−5x2+10x+15h(x)=-5x^{2}+10x+15\newlineHow many seconds after being thrown will the stone hit the water?\newlineseconds\text{seconds}

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Q. Alain throws a stone off a bridge into a river below.\newlineThe stone's height (in meters above the water), xx seconds after Alain threw it, is modeled by:\newlineh(x)=−5x2+10x+15h(x)=-5x^{2}+10x+15\newlineHow many seconds after being thrown will the stone hit the water?\newlineseconds\text{seconds}
  1. Write Equation for Stone's Height: Write down the equation that models the stone's height above the water.\newlineThe equation given is h(x)=−5x2+10x+15h(x) = -5x^2 + 10x + 15, where h(x)h(x) is the height in meters and xx is the time in seconds after the stone is thrown.
  2. Set Height to 00: Set the height h(x)h(x) to 00 to find when the stone will hit the water.\newline0=−5x2+10x+150 = -5x^2 + 10x + 15
  3. Factor Quadratic Equation: Factor the quadratic equation to solve for xx.\newlineTo factor, we look for two numbers that multiply to −5×15=−75-5 \times 15 = -75 and add up to 1010. However, since the quadratic is not easily factorable, we will use the quadratic formula instead.
  4. Apply Quadratic Formula: Apply the quadratic formula to find the values of xx.\newlineThe quadratic formula is x=−b±b2−4ac2ax = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{{2a}}, where a=−5a = -5, b=10b = 10, and c=15c = 15.
  5. Calculate Discriminant: Calculate the discriminant b2−4acb^2 - 4ac.\newlineDiscriminant = 102−4(−5)(15)=100−(−300)=100+300=40010^2 - 4(-5)(15) = 100 - (-300) = 100 + 300 = 400
  6. Calculate Possible Values for x: Calculate the two possible values for x using the quadratic formula.\newlinex=−10±4002⋅−5x = \frac{{-10 \pm \sqrt{400}}}{{2 \cdot -5}}\newlinex=−10±20−10x = \frac{{-10 \pm 20}}{{-10}}
  7. Solve for x: Solve for the two possible values of x.\newlinex=−10+20−10=10−10=−1x = \frac{{-10 + 20}}{{-10}} = \frac{{10}}{{-10}} = -1 (This value is not physically meaningful as time cannot be negative.)\newlinex=−10−20−10=−30−10=3x = \frac{{-10 - 20}}{{-10}} = \frac{{-30}}{{-10}} = 3 (This is the time in seconds when the stone hits the water.)

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