I'm asked to calculate a stone being dropped from rest to hit the water it takes 2.7s I found the final velocity to be 26.46m/s

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    $\begingroup$ I believe the way you are doing it you have two unknown variables. The initial and final velocity will be different if you threw the stone down. If you know the height, you can use the kinematic equation to solve for the initial velocity. $\endgroup$ – Tyberius Jun 14 at 1:43
  • $\begingroup$ $x_f=x_i+v_it+\frac{1}{2}at^2$ $\endgroup$ – Tyberius Jun 16 at 21:37

No, because you are assuming that the stone hits the water at the same speed in both cases. If you fire a rifle bullet into the water then you will see that that cannot be true. (Good general principle in thought experiments: before getting seduced by numbers and formulae, make a picture in your mind of an extreme case and see how the concepts fit).

The stone will hit the water at a speed $2.3g$ greater at that with which you threw it. The average speed of the stone will therefore be $1.15g$ plus your throw. The distance the stone fell will therefore be $2.3$ times that.

Since, from dropping the stone and it taking $2.7$ seconds you know what the distance to the water is, you can work backwards and get the speed of the throw.


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