To solve for t, use the formula:

`s=v_it + 1/2at^2`

where `s` - displacement `v_i` - initial velocity

`t` - time `a` - accelaration

Note that displacement refers to the change in height. So value of s is:

`s=h_f-h_i=0-3.82=-3.82`

Also, acceleration is equal to gravity since the ball exhibits the principle of freely falling bodies. Its sign is negative because the direction of the intial velocity is upward. So, `a=-9.81 m/s^2` .

Then, substitute `s=-3.82` , `v_i=1.86` and `a=-9.82` .

`-3.82=1.86t + 1/2(-9.82)t^2`

`-3.82=1.86t-4.905t^2`

Express the equation in quadratic form `ax^2+bx+c=0` .

`4.905t^2-1.86t-3.82=0`

Apply the quadratic formula to solve for t.

`t=[-b+-sqrt(b^2-4ac)]/(2a)=[-(-1.86)+-sqrt((-1.86)^2-4(4.905)(-3.82))]/(2(4.905))`

`t=[1.86+-sqrt78.408]/9.81=(1.86+-8.855)/9.81`

`t_1=1.09` and `t_2= -0.71`

Since t represents time, take only the positve value.

**Hence, the time it takes for the ball to reach the ground is 1.09 seconds.**

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