The body is thrown vertically upward. What are the initial speed and time of movement

The body is thrown vertically upward. What are the initial speed and time of movement if the body has visited the height h twice with an interval of t0?

Given:

g is the acceleration of gravity;

h – some height;

t0 is the time interval during which the body has visited the h mark for 2 times.

It is required to determine the initial velocity of the body v0 and the total flight time t.

According to the condition of the problem, the body was thrown vertically upward. We also do not take into account the resistance of the medium. Then, according to the law of conservation of energy, the sum of potential and kinetic energy at any moment of the body’s flight are the same. Let’s find this sum at the height h. According to the condition of the problem, the body has visited this height 2 times. This means that, having reached the height h, it continued its flight to the maximum height Hmax, and then returned again in the process of falling to the height H. Let the speed of the body at the height h be equal to v. It is known that at maximum altitude the speed of a body is zero. Then:

v – g * t0 / 2 = 0;

v = g * t0 / 2, where t0 / 2 is the time interval during which the body flew from point h to point Hmax.

Then, according to the law of conservation of energy:

m * g * h + m * g ^ 2 * t0 ^ 2/8 = m * v0 ^ 2/2;

g * h + g ^ 2 * t0 ^ 2/8 = v0 ^ 2/2;

8 * g * h + g ^ 2 * t0 ^ 2 = 4 * v0 ^ 2;

v0 ^ 2 = 2 * g * h + g ^ 2 * t0 ^ 2/4;

v0 ^ 2 = (4 * g * h + g ^ 2 * t0 ^ 2) / 4;

v0 = ((4 * g * h + g ^ 2 * t0 ^ 2) / 4) ^ 0.5 = (4 * g * h + g ^ 2 * t0 ^ 2) ^ 0.5 / 2.

Then the total flight time will be:

t = 2 * v0 / g = (4 * g * h + g ^ 2 * t0 ^ 2) ^ 0.5 / g.

Answer: the initial velocity of the body is (4 * g * h + g ^ 2 * t0 ^ 2) ^ 0.5 / 2, the total flight time of the body is (4 * g * h + g ^ 2 * t0 ^ 2) ^ 0 , 5 / g.