Find the height and speed of an arrow released vertically upward with an initial speed of 50 m / s in 4 s.

The arrow moves equally slowly and the formula for the distance traveled is:
S = (V² – V0²) / 2a, where V is the final speed, V0 is the initial speed (V0 = 50 m / s), and is the acceleration (a = -g (g = 10 m / s²)).
Boom speed:
V = V0 + a * t, where t is the time of movement (t = 4 s).
-g * t = V – V0.
V = V0 – g * t = 50 – 10 * 4 = 10 m / s.
Let’s calculate the distance traveled:
S = (V² – V0²) / (-2g) = (10² – 50²) / (-2 * 10) = -2400 / (-20) = 120 m.
Answer: The lifting height is 120 m, the boom speed is 10 m / s.