A stone is thrown up from the tower, telling it a speed of 10 m / s where will the stone

A stone is thrown up from the tower, telling it a speed of 10 m / s where will the stone be in 2s if the height of the tower is 15 m, when will it fall to the ground?

Given: V1 (throwing speed) = 10 m / s; t (considered travel time) = 2 s; hb (tower height) = 15 m; g ≈ 10 m / s2.

1) The duration of the rise of the stone to the maximum height: 0 = V0 – g * t1 and t1 = V0 / g = 10/10 = 1 s; therefore, after another 1 s (2 s after the start of movement) the stone will be at the level of the top of the tower …

2) Lift height: h = V1 ^ 2 / 2g = 10 ^ 2 / (2 * 10) = 5 m.

3) The time the stone fell to the ground from the moment of throwing: tp = t1 + t2 = t1 + √ (2 * (hb + h) / g) = 1 + √ (2 * (15 + 5) / 10) = 3 s.