Answer to Question #230353 in Civil and Environmental Engineering for syra

Question

A stone is thrown downward with an initial velocity of 5m/s from a 20-meter high building. Find the time it takes to reach the ground.

Answer 1

$h= ut + \frac{1}{2}gt^2\\ 20= 5t + \frac{1}{2}*9.8*t^2\\ \mathrm{Multiply\:both\:sides\:by\:}10\\ 20\cdot \:10=5t\cdot \:10+\frac{1}{2}\cdot \:9.8t^2\cdot \:10\\ 50t+49t^2=200\\ \mathrm{Subtract\:}200\mathrm{\:from\:both\:sides}\\ 50t+49t^2-200=200-200\\ \mathrm{Simplify}\\ 49t^2+50t-200=0\\ t_{1,\:2}=\frac{-50\pm \sqrt{50^2-4\cdot \:49\left(-200\right)}}{2\cdot \:49}\\ t_1=\frac{-50+10\sqrt{417}}{2\cdot \:49},\:t_2=\frac{-50-10\sqrt{417}}{2\cdot \:49}\\ t=\frac{5\left(\sqrt{417}-5\right)}{49},\:t=-\frac{5\left(5+\sqrt{417}\right)}{49}\\ t=1.57352 s$