Question

A stone thrown from ground level returns to the same level 4 s after. With what speed was the stone thrown? Take g=10ms−2

Answer 1

The motion is symmetrical vertically - so after 2 s it will reach the top of its path, and have a vertical speed of 0

v =ut +0.5 a t^2

0 = 2*u + 5*2^2

so calculate u = initial VERTICAL speed.

The question does not state that the stone was thrown vertically - BUT you cannot calculate anything about horizontal speed from the information given , so you have to assume the motion is straight up and down.