Question

A ball is thrown vertically upward with a speed of 1.86

m/sfrom a point 3.82 m above the ground. calculate the time in
whichthe ball will reach the ground.

Answer 1

t = 1.09 s.

Step-by-step explanation:

This is a one-dimensional kinematics question, so the equations of kinematics will be sufficient to solve the question.

`y-y_0 = v_0t + (1)/(2)at^2\\0 - 3.82 = 1.86t +(1)/(2)(-9.8)t^2\\-3.82 = 1.86t - (1)/(2)9.8t^2\\4.9t^2 - 1.86t - 3.82 = 0`

This quadratic equation can be solved using determinant.

`\Delta = b^2 - 4ac\\t_(1,2) = (-b \pm √(\Delta) )/(2a)\\t_1 = 1.09~s\\t_2 = -0.71~s`

Of course, we will choose the positive time.