# The body is thrown at an angle from a height of 10 m above the ground at a speed of 20

**The body is thrown at an angle from a height of 10 m above the ground at a speed of 20 m / s. What will be its speed at a height of 25 m?**

To determine the value of the speed of a thrown body at a height of 25, we apply the law of conservation of energy: Enn + Ekn = Epk + Ekk and m * g * hn + m * Vn ^ 2/2 = m * g * hk + m * Vk ^ 2/2 , whence we express: Vk2 / 2 = g * hn + Vn ^ 2/2 – g * hk and Vk = √ (2 * (g * hn + Vn ^ 2/2 – g * hk)).

Constants and variables: g – gravitational acceleration (g ≈ 10 m / s2); hн – initial body height (hн = 10 m); Vн – throw speed (Vн = 20 m / s); hk – considered (final) height (hk = 25 m).

Calculation: Vк = √ (2 * (g * hн + Vн ^ 2/2 – g * hк)) = √ (2 * (10 * 10 + 20 ^ 2/2 – 10 * 25)) = 10 m / s …

Answer: At a height of 25 m, the speed of the body was 10 m / s.