Question

Kevin threw a ball straight up with an initial speed of 20 meters per second.

The function h(t) = –5(x — 2)2 + 20 describes the ball’s height, in meters, t seconds after Kevin threw it. What is the maximum height of the ball?

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A)
2 meters

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B)
5 meters

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C)
10 meters

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D)
20 meters

Answer 1

Answer:

it should be D) 20 meters, not 100% sure tho

Answer 2

From the vertex of the quadratic equation, it is found that the maximum height of the ball is of 20 meters, given by option D.

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

$h(t) = -5(x^2 - 4x + 4) + 20$

$h(t) = -5x^2 + 20x - 20 + 20$

$h(t) = -5^x^2 + 20x$

Which is a quadratic equation with coefficients $a = -5, b = 20, c = 0$.

It is a concave down parabola, thus, it's maximum value is the h-coordinate of the vertex, which is:

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

Then, inserting the values:

$h_V = -\frac{20^2 - 4(-5)(0)}{4(-5)} = 20$

The maximum height of the ball is of 20 meters, given by option D.

A similar problem is given at https://brainly.com/question/24626341