The body is thrown up at a speed of 20 m / s. How long after the throw will it reach a height of 15 m?

V0 = 20 m / s.
g = 10 m / s2.

h = 15 m.

t -?

The body when moving upwards is decelerated with the acceleration of gravity g.

t = (V0 – V) / g.

We find the speed of the body at the height h from the law of conservation of total mechanical energy.

m * V0 ^ 2/2 = m * V ^ 2/2 + m * g * h.

V = √ (V0 ^ 2 – 2 * g * h).

V = √ ((20 m / s) ^ 2 – 2 * 10 m / s2 * 15 m) = 10 m / s.

t = (20 m / s – 10 m / s) / 10 m / s2 = 1 s.

Answer: a body at a height of h = 15 m will be in time t = 1 s.