The equation y = -16t^2 + 28t + 180 describes the height (in feet) of a ball tossed up in the air at 28 feet per second from a height of 180 feet from the ground. In how many seconds will the ball hit the ground?
+14991 In how many seconds will the ball hit the ground?
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\(h=v_0t-\dfrac{t^2}{2}\cdot g\\ -180ft=\frac{28ft}{s}t-\dfrac{t^2}{2}\cdot \dfrac{9.81\ m}{s^2}\cdot \dfrac{3.28\ ft}{m}\ |\ :\ ft\\ -180=\dfrac{28}{s}t-t^2\cdot \dfrac{9.81}{s^2}\cdot\ 1.64\\ t^2\cdot \dfrac{9.81}{s^2}\cdot\ 1.64-\dfrac{28}{s}t-180=0\)
\(t = {\dfrac{28}{s} \pm \sqrt{\dfrac{784}{s^2} +4\cdot \dfrac{9.81\cdot 1.64}{s^2}\cdot 180} \over 2\cdot \dfrac{9.81}{s^2}\cdot\ 1.64}\\ t= \dfrac{28s\pm \sqrt{784+4\cdot 9.81\cdot 1.64\cdot 180}\ s}{2\cdot9.81\cdot 1.64}\\ \color{blue}t=4.215\ s\)
In 4.215 seconds will the ball hit the ground.
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