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A waterfall has a height of 900 feet. A pebble is thrown upward from the top of the falls with an initial velocity of 20 feet per second. The height, [tex]h[/tex], of the pebble after [tex]t[/tex] seconds is given by the equation:

\[ h = -16t^2 + 20t + 900 \]

How long after the pebble is thrown will it hit the ground?

The pebble will hit the ground about _____ seconds after it is thrown. (Simplify your answer. Round to one decimal place as needed.)

Answer :

Final answer:

The pebble thrown from the top of a 900 foot waterfall with an initial upward velocity of 20 feet per second will hit the ground about 45 seconds after it is thrown, calculated using the quadratic formula.

Explanation:

To find the time when the pebble hits the ground, we set the height equation h = -16t^2 + 20t + 900 to zero (since once the pebble hits ground, the height will be zero) and solve for t. This turns into a quadratic equation, which can be solved using the quadratic formula: t = [-b ± sqrt(b^2 - 4ac)] / 2a. In this equation, a = -16, b = 20, and c = 900.

Substitute these values into the quadratic formula to solve for t. We get two solutions, t1 = -1.25 and t2 = 45. Remember that time cannot be negative, so we discard the first solution. The answer is t = 45 seconds. So, the pebble will hit the ground approximately 45 seconds after it is thrown.

Learn more about quadratic equation here:

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