Answer:
t = 1.09 s.
Explanation:
This is a one-dimensional kinematics question, so the equations of kinematics will be sufficient to solve the question.
[tex]y-y_0 = v_0t + \frac{1}{2}at^2\\0 - 3.82 = 1.86t +\frac{1}{2}(-9.8)t^2\\-3.82 = 1.86t - \frac{1}{2}9.8t^2\\4.9t^2 - 1.86t - 3.82 = 0[/tex]
This quadratic equation can be solved using determinant.
[tex]\Delta = b^2 - 4ac\\t_{1,2} = \frac{-b \pm \sqrt{\Delta} }{2a}\\t_1 = 1.09~s\\t_2 = -0.71~s[/tex]
Of course, we will choose the positive time.
