Hello!

After an object was thrown, the only force acting on it is the gravity force (if we ignore air resistance). It acts downwards and is constant near Earth's surface. For an object of mass `m` the gravity force is `mg,` where `g=9.8 m/s^2` is the gravity acceleration.

By the Newton's Second law a constant force `F` gives a body of mass `m` a constant acceleration `F/m,` in our case it is `g.` Constant acceleration `g` means that the velocity will be

`V_0-g t`

(minus because `V_0` has the opposite direction to `g` ),

and the height will be

`V_0t-g (t^2)/2.`

An object hits ground when the height is zero and `t gt 0:`

`V_0t-g (t^2)/2=0,` or

`V_0-g (t)/2=0,` or

`t=(2V_0)/(g)=2(s).`

So the answer is: an object will hit the ground after** 2** seconds.

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