The balloon rises evenly vertically upward at a speed of 4 m / s. A load is suspended

The balloon rises evenly vertically upward at a speed of 4 m / s. A load is suspended from it on a rope. At an altitude of 217m, the rope breaks. How many seconds will the load fall to the ground if g = 10 m / s²

V = 4 m / s.

h = 217 m.

g = 10 m / s2.

t -?

Since the load is attached by a rope to the balloon, which rises upward at a speed V, the load at the moment of separation will have a speed V directed vertically upward.

Only the force of gravity acts on the load, so it will move with the acceleration of free fall g, first upward until it comes to a complete stop, and then downward.

t = t1 + t2.

t1 = V / g.

t1 = 4 m / s / 10 m / s2 = 0.4 s.

The height of the rise is expressed by the formula: h1 = V2 / 2 * g.

h1 = V2 / 2 * g.

h1 = (4 m / s) 2/2 * 10 m / s2 = 0.8 m.

t2 = √ (2 * (h1 + h) / g).

t2 = √ (2 * (0.8 m + 217 m) / 10 m / s2)) = 6.6 s.

t = 0.4 s + 6.6 s = 7 s.

Answer: t = 7 s.