Mathematics: Using a quadratic equation A ball is thrown from an initial height of 1 meter with an initial upward velocity of 20 m/s. The ball s height h (in meters) after 1 seconds is given by the following h=1+20t-5t2 Find all values of 1 for which the ball s height is 11 meters. Round your answer(s) to the nearest hundredth. (If there is more than one answer or )

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Using a quadratic equation A ball is thrown from an

gabrielaaaa1323

22.04.2021

The ball height is 11 meters at t = 0.59 seconds and t = 3.41 seconds.

Step-by-step explanation:

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:

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

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

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

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

h(t) = 1 + 20t - 5t^2

Find all values of t for which the ball's height is 11 meters.

This is t for which h(t) = 11 . So

h(t) = 1 + 20t - 5t^2

11 = 1 + 20t - 5t^2

5t^2 - 20t + 10 = 0

Simplifying by 5

t^2 - 4t + 2 = 0

Which means that a = 1, b = -4, c = 2

$\Delta = (-4)^2 - 4(1)(2) = 8$

$t_{1} = \frac{-(-4) + \sqrt{8}}{2} = 3.41$

$t_{2} = \frac{-(-4) - \sqrt{8}}{2} = 0.59$

The ball height is 11 meters at t = 0.59 seconds and t = 3.41 seconds.

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