Answer: [tex]\begin{gathered} Quadratic\text{ equation: }4.9t^2\text{ - 9.8t - 73.5} \\ The\text{ time it will take the cannonball to reach sea level is 5 seconds} \end{gathered}[/tex]
Explanation:
Given:
distance from sea level to top of hill = initial heeight = 73.5 meters
velocity = 9.8 m/s
[tex]\begin{gathered} For\text{ vertical movement:} \\ Final\text{ height = acceleration\lparen t}^2)\text{ + velocity\lparen t\rparen+ initial height} \\ Since\text{ it is reaching sea level, final height = 0} \\ acceleration\text{ = -9.8 m/s}^2 \\ \\ 0\text{ = -}\frac{1}{2}(9.8)t^2\text{ + 9.8t + 73.5m} \end{gathered}[/tex][tex]\begin{gathered} 0\text{ = -4.9t}^2\text{ + 9.8t + 73.5} \\ 4.9t^2\text{ - 9.8t - 73.5 = 0 \lparen quadratic equation\rparen} \\ \\ \text{Using formula method to find the value of t:} \\ t\text{ = }\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ t\text{ = }\frac{-(-9.8)\pm\sqrt{(-9.8)^2-4(4.9)(-73.5)}}{2(4.9)} \\ \text{ t = }\frac{-(-9.8)\pm\sqrt{1536.64}}{9.8} \\ t\text{ = }\frac{9.8\pm39.2}{9.8} \end{gathered}[/tex][tex]\begin{gathered} t\text{ = }\frac{9.8+39.2}{9.8}\text{ ot }\frac{9.8\text{ - 39.2}}{9.8} \\ \\ t\text{ = 5 or -3} \end{gathered}[/tex]
Since we can't have t to be negative, t = 5
The cannonball will reach the sea level at 5 seconds
