The body was thrown vertically upward at a speed of v = 30 m / s.

The body was thrown vertically upward at a speed of v = 30 m / s. Find its speed at the moment it was at half its maximum height for the first time.

Given:

V0 = 30 m / s – the speed with which the body was thrown;

g = 10 m / s ^ 2 – acceleration of gravity.

It is required to find V1 – the speed at the moment when the body was at half the maximum height.

First, let’s find the height:

S = V0 ^ 2/2 * g = 30 * 30/2 * 10 = 900/20 = 45 meters, that is, half the height will be h = 45/2 = 22.5 m.

Further, according to the law of conservation of energy:

m * V0 ^ 2/2 = m * V1 ^ 2/2 + m * g * h or

V0 ^ 2 = V1 ^ 2 + 2 * g * h, hence:

V1 = square root (V0 ^ 2 – 2 * g * h) = square root (30 ^ 2 – 2 * 10 * 22.5) =

= square root (900 – 450) = square root (450) = 21.2 m / s.

Answer: the speed at half maximum height will be 21.2 m / s.