Posted by Julie on Saturday, March 6, 2010 at 6:15pm.
The same speed that it was thrown, but travelling in the opposite direction. You can conclude that from an energy conservation argument.
The kinetic energy when thrown and landing equals the potential energy at the highest point of the trajectory.
(1/2) M V^2 = M g H
The time of flight T satisfies
H = (V/2)T
where T is the time of flight.
Thus V^2 = 2 g H = g T V
V = g T
There is an easier way to reach the same result, but I didn't see it.
T = V/g is the time it takes for the rock to reach zero velocity, which it does at the top of the trajectory.
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