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 24 feet per second. the height, h, of the pebble after t seconds is given by the equation h = - 16t² + 24t + 1500. 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.)

Explanation:

Step1: Set height h = 0

We want to find when the pebble hits the ground, so we set the height - equation $h=-16t^{2}+24t + 1500$ equal to 0. So, $-16t^{2}+24t + 1500=0$. Divide through by - 8 to simplify: $2t^{2}-3t - 187.5=0$.

Step2: Use the quadratic formula

The quadratic formula for a quadratic equation $ax^{2}+bx + c = 0$ is $t=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$. For the equation $2t^{2}-3t - 187.5=0$, we have $a = 2$, $b=-3$, and $c=-187.5$. First, calculate the discriminant $\Delta=b^{2}-4ac=(-3)^{2}-4\times2\times(-187.5)=9 + 1500=1509$.

Step3: Find the values of t

$t=\frac{-(-3)\pm\sqrt{1509}}{2\times2}=\frac{3\pm\sqrt{1509}}{4}$. We have two solutions for t: $t=\frac{3+\sqrt{1509}}{4}$ and $t=\frac{3 - \sqrt{1509}}{4}$. Since time cannot be negative, we discard the solution $t=\frac{3-\sqrt{1509}}{4}$. $\sqrt{1509}\approx38.85$. Then $t=\frac{3 + 38.85}{4}=\frac{41.85}{4}=10.4625\approx10.5$.

Answer:

10.5