The boy dives into the water from a 3m high steep bank with an initial speed of 7m / s.

The boy dives into the water from a 3m high steep bank with an initial speed of 7m / s. Determine the speed of the boy when he reaches water.

Initial data: h (steep bank height) = 3 m; V1 (boy’s initial speed) = 7 m / s.

Reference values: g (acceleration due to gravity) ≈ 10 m / s2 = a (acceleration of the boy).

We express the boy’s speed when he reaches water from the equality (the law of conservation of energy): Ek2 = Ek1 + En1.

m * V2 ^ 2/2 = m * V1 ^ 2/2 + m * g * h.

V2 = √ (V1 ^ 2 + 2g * h).

Calculation: V2 = √ (7 ^ 2 + 2 * 10 * 3) = √109 = 10.44 m / s.

Answer: Having reached the surface of the water, the boy will have a speed of 10.44 m / s.