# Free falls

What 's the difference between an object thrown or dropped , does a thrown object treat as a free fall condition ?? if it does
we know that d=v(initial)*t + 1/2a*t^2
and we know that in free falls condition we use the equation d= 1/2g*t^2
that means initial velocity of free falls should be zero and thrown object has an initial velocity ??
can any one explain to me this , please >>>???

It's hard to make out what you are asking but I think that what you are saying is correct.

If we release (not throw), a ball for instance, on earth relative to earth then it's initial velocity is fairly obviously zero. You were just holding the ball before it dropped, it's was not moving relative to you or the earth. It clearly doesn't have any velocity (it's not moving!)

Now if we throw a ball we swing our arm before the ball is released. The ball is moving with our arm but the ball is still moving so when it it released the ball will have some initial velocity

EDIT: In both instances there is the same acceleration (9.8 m/s^2) the only difference is that in the throwing scenario we are giving the ball an starting velocity.

$$x(t) = x_0 + v_0 t + \frac{g t^2}{2}$$ will work in both cases, but if we drop a ball then $$v_0=0$$ in that equation.

Yes...in everyday speaking the difference between 'throwing' and 'dropping' (or 'lifting') is the initial velocity.The first one implies that the initial velocity is not zero while the second implies that it is.