A body weighing 1 kilogram is thrown vertically upward at a speed of 10 meters per second, absolutely resiliently

A body weighing 1 kilogram is thrown vertically upward at a speed of 10 meters per second, absolutely resiliently hits a horizontal obstacle located at a height of 1.8 m and bounces off it. Determine the impulse of the impact force (N * s)

Given:
m = 1kg,
v0 = 10m / s,
h = 1.8m;
Find: F * t -?
With an absolutely elastic impact, the body bounces off the obstacle with the same velocity (in magnitude) with which it struck, but directed in the opposite direction.
The impulse of force is equal to the change in the impulse of the body:
F * t = Δp = m * Δv = m * v – m * (-v) = 2 * m * v;
Here v is the speed of the body at height h. Let’s find her.
With uniformly accelerated motion of the body, the path traveled by it is equal to:
h = (v0 ^ 2 – v ^ 2) / (2 * g),
where g = 10m / s ^ 2 is the acceleration of gravity;
From here we find v:
v ^ 2 = v0 ^ 2 – 2 * g * h = 100 – 36 = 64;
v = 8m / s;
F * t = 2 * 1 * 8 = 16H * s.