A stone thrown vertically upward returned to the ground after t = 4 s. Find the initial speed of the stone.

g = 9.8 m / s2.

t = 4 s.

V0 -?

When moving, only gravity acts on the stone, so it moves with the acceleration of gravity g. Up the stone moves equally slowed down to a complete stop, and downward uniformly accelerated.

t = t1 + t2, where t1 is the time the stone moves up, t2 is the time the stone moves down.

Since when moving up, the final speed of the stone is V = 0, and when moving down, the initial speed of the stone is 0 and they move with the same acceleration, then t1 = t2 = t / 2.

g = (V0 – V) / t1 = 2 * V0 / t.

The initial speed of the stone will be expressed by the formula: V0 = g * t / 2.

V0 = 9.8 m / s2 * 4 s / 2 = 19.6 m / s.

Answer: at the moment of throwing the stone had a speed V0 = 19.6 m / s.