A ball fired vertically upward from a spring pistol covered a distance of 30 m in some time.

A ball fired vertically upward from a spring pistol covered a distance of 30 m in some time. The modulus of its movement during the same time was 6 m. What was the initial velocity of the ball?

Since at the moment in question, the modulus of movement of the ball is 6 m (distance to the launch point), and during the same time it has already covered a path of 30 m, then the ball will have to travel 30 + 6 = 36 (from launch to return to the starting point) m). Of these, 36: 2 = 18 (m) up and the same amount down, that is, the height of the ball is 18 m. To calculate the height of the flight of a body thrown vertically up, we take the formula: H = (Δv) ^ 2: (2 g). At the top of the flight, the speed of the ball is zero, so H = (v) ^ 2: (2 g); (v) ^ 2 = 2 g H; (v) ^ 2 = 2 * 9.8 * 18, (v) ^ 2 = 352.8; we extract the root and find the initial velocity of the ball: v = 18.8 (m / s).
Answer: v = 18.8 m / s.