A ball is thrown directly upward from a height of 8 ft with an initial velocity of 24 ft/sec. The function s(t) = −16t²+24t+8 gives the height of the ball, in feet, t seconds after it has been thrown. a. Determine the time at which the ball reaches its maximum height and find the maximum height. b. How long after the ball is thrown does it land back on the ground?

(a)0.625s (b)1.569s

Step-by-step explanation:

it's Physics, not Math.

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With this algebraic representation of physics, we can also apply the mathmatical concept of quadratics to derive the components from this function. You can use the line of symmetry: x = -b/2a. To find the minimum or maximum height because of its association with the vertex. This function is in the standard quadratic form: ax^2 + bx + c where x is t. According to this instance, a being -16, b being 24, and c being 8. t = -b/2a → -24/-32 = 3/4 which also represents the maximum value because a is negative so it is opening downward.

To find the max height, substitute the value of x back into the equation f(3/4) = -16(3/4)^2 + 24(3/4) + 8 →

-9 + 18 + 8 = 17ft [maximum height]

We already solved for the time (x) in terms of t which is 3/4 sec [time of maximum height].

make sure you have version 2.50

try looking in the very back of the book : )

step-by-step explanation:

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