The balloon at rest begins to rise from the ground with an acceleration of 2 m / s2.

The balloon at rest begins to rise from the ground with an acceleration of 2 m / s2. In 5 s after the start of the ascent, the ballast is dropped from it (without the initial speed relative to the balloon). How soon will the ballast fall to the ground?

Given:
V0 = 0 is the initial velocity of the ball,
a = 2m / s ^ 2,
t1 = 5s,
h = 0
Find: t2 -?
Find the speed of the ball
V = a * t1 = 10m / s
and height after 5s ascent:
H = 1/2 * a * (t1) ^ 2 = 25m;
Let’s write the equations of motion for the ballast:
h = H + u0 * t – 1/2 * g * t ^ 2;
Here
H – the height of the ballast at the beginning of the movement,
u0 = V – initial velocity directed upwards,
g = 10m / s ^ 2 – free fall acceleration, taken with a minus sign, since it is directed downward,
and it is necessary to find the time t2 when the ballast height h = 0;
H + u0 * t2 – 1/2 * g * (t2) ^ 2/2 = 0;
5 * (t2) ^ 2 – 10 * t2 – 25 = 0;
(t2) ^ 2 – 2 * t2 – 5 = 0;
t2 = 1 ± sqrt (1 + 5);
t2 = 3.45s.