Question

The height of a cannon ball, in feet, shot from a pirate ship is given by
the equation h = -0.03x* + 2.84x + 20, where x is given in seconds
After how many seconds does the cannon ball splash into the water??

Answer 1

Answer:

after 101.25 seconds

Step-by-step explanation:

set h equal to zero and solve for 'x'

0 = -0.03x² + 2.84x + 20

I multiplied through by 100 to work with integers instead of decimals, and then I used the Quadratic Formula

9-b ± √b²-4ac) / 2a

a = -3

b = 284

c = 2000

Answer 2

Solving the quadratic equation, it is found that the ball splashes into the water after 101.25 seconds.

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The height of the ball after x seconds is given by:

$h(x) = -0.03x^2 + 2.84x + 20$

• It splashes into the water at time x, for which $h(x) = 0$. This value is found solving the quadratic equation.

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Solving a quadratic equation:

Given a second order polynomial expressed by the following equation:

$ax^{2} + bx + c, a\neq0$.

This polynomial has roots $x_{1}, x_{2}$ such that $ax^{2} + bx + c = a(x - x_{1})*(x - x_{2})$, given by the following formulas:

$x_{1} = \frac{-b + \sqrt{\Delta}}{2*a}$

$x_{2} = \frac{-b - \sqrt{\Delta}}{2*a}$

$\Delta = b^{2} - 4ac$

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$-0.03x^2 + 2.84x + 20$ has coefficients $a = -0.03, b = 2.84, c = 20$. Thus:

$\Delta = (2.84)^2 - 4(-0.03)(20) = 10.4656$

$x_{1} = \frac{-2.84 + \sqrt{10.4656}}{2(-0.03)} = -6.58$

$x_{2} = \frac{-2.84 - \sqrt{10.4656}}{2(-0.03)} = 101.25$

The ball splashes into the water after 101.25 seconds.

A similar problem is given at https://brainly.com/question/16840613