Question

A waterfall has a height of 1500 feet. A pebble is thrown upward from the top of the falls with an initial velocity of 20 feet per second. The height, h, of the pebble after t seconds is given by the equation h=-16t+20t+1500. How long after the pebble is thrown will it hit the ground?

Answer 1

If you can copy the equation correctly, a graphing calculator can tell you the height (h) will be zero when t is about 10.328.

The pebble will hit the ground (h=0) about 10.328 seconds after it is thrown.

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You want to find t that makes h=0, so you are looking for solutions to

$0=-16t^2+20t+1500\\0=4t^2-5t-375\qquad\text{divide by -4}\\t=\dfrac{5\pm\sqrt{(-5)^2-4\cdot4\cdot(-375)}}{2\cdot4}\qquad\text{using the quadratic formula}\\t=\dfrac{5\pm\sqrt{6025}}{8}=0.625+\sqrt{94.140625}\qquad\text{for t {data-answer}gt; 0}\\t\approx 10.3276$