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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 [tex]h = -16t^2 + 20t + 900[/tex].

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

Answer :

Final answer:

To find the time when a pebble hits the ground after being thrown from a waterfall (given that the height, initial velocity and gravity form a quadratic equation), we set h to 0 and solve the quadratic equations using the quadratic formula. The positive solution will be the time duration before the pebble hits the ground.

Explanation:

The subject of this question is quadratic physics, split into projectile motion and solving quadratic equations. In this case, the pebble is thrown from a height (h), initially upwards with a certain velocity, and it eventually falls downward due to gravity. The motion of the pebble can be mathematically modeled using a quadratic equation, reflecting the parabolic trajectory the pebble follows in its flight. In the described situation, the pebble will fall to the ground when h=0.

To solve this problem, we will set h to 0 in the given equation, creating the quadratic equation -16t²+20t+900 = 0. We then have to solve this equation to find the values for 't' when h=0. We can solve a quadratic equation in the form at²+bt+c =0 using the quadratic formula: t = [-b ± sqrt(b² - 4ac)] / (2a).

Filling in the parameters of the quadratic equation into the formula, we find: t = [-20 ± sqrt((20)² - 4*(-16)*900)] / (2*(-16)). Solving this equation gives two results, one of which will be a negative number representing the time before the pebble was thrown, which we discard leaving us with the time t = n seconds where n is the positive value we found. This will be the time duration before the pebble hits the ground.

Learn more about Solving Quadratics here:

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