The balloon flies at an altitude of 2 km at a speed of 10 m / s. Ballast is dropped from the balloon.

The balloon flies at an altitude of 2 km at a speed of 10 m / s. Ballast is dropped from the balloon. Determine the time and distance of the ballast drop.

Vhor = 10 m / s.
g = 9.8 m / s ^ 2.
h = 2 km = 2000 m.
V0 = 0 m / s.
t -?
L -?
Ballast vertically will fall freely with gravitational acceleration g = 9.8 m / s ^ 2 without an initial velocity.
h = g * t ^ 2/2.
t = √ (2 * h / g).
t = √ (2 * 2000 m / 9.8 m / s ^ 2) = 20.2 s.
Horizontally, the balloon will move away from the ballast evenly at a speed Vhor.
For uniform movement, the distance is found by the formula: L = Vhor * t.
L = 10 m / s * 20.2 s = 202 m.
Answer: the time of the ballast fall is t = 20.2 s, the fall distance is L = 202 m.