A waterfall has a height of 1100 feet. A pebble is thrown upward from the top of the falls with an initial velocity of 20 feet per second. The height of the pebble h in feet after t seconds is given by the equation h = -16 t^2 + 20*t + 1100. How long after the pebble is thrown will it hit the​ ground?

Answer:8.94 secondsStep-by-step explanation:The ground is at height 0.We set the equation equal to zero and solve for t.h = -16 t^2 + 20*t + 1100We want h = 0, so we get-16 t^2 + 20*t + 1100 = 04t^2 - 5t - 275 = 0[tex] t = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} [/tex][tex] t = \dfrac{-(-5) \pm \sqrt{(-5)^2 - 4(4)(-275)}}{2(4)} [/tex][tex] t = \dfrac{5 \pm \sqrt{25 + 4400}}{8} [/tex][tex] t = \dfrac{5 \pm \sqrt{4425}}{8} [/tex][tex] t = \dfrac{5 \pm 5\sqrt{177}}{8} [/tex][tex] t = 8.94 [\tex]   or   [/tex] t = -7.69 [/tex]We discard the negative answer.Answer: 8.94 seconds