At the bow of a 5 m long boat, a man stands holding a stone weighing 4 kg at a height of 1.25 m above the boat.

At the bow of a 5 m long boat, a man stands holding a stone weighing 4 kg at a height of 1.25 m above the boat. A man throws a stone horizontally along the boat. What speed relative to the ground must a person communicate to a stone in order to get into the stern of the boat? The mass of the boat with a person is 246 kg, do not take into account the resistance of water and air. Take the acceleration due to gravity equal to 10 m / s2.

L = 5 m.
h = 1.25 m.
mk = 4 kg.
ml = 246 kg.
g = 10 m / s ^ 2.
V -?
The movements of the stone can be divided into two types: vertically, the stone will move with acceleration g, horizontally evenly in a straight line.
Let’s find the free fall time of a stone t from a height of h = 1.25 m.
h = g * t ^ 2/2.
t = √ (2 * h / g).
t = √ (2 * 1.25 m / 10 m / s ^ 2) = 0.5 s.
According to the law of conservation of momentum, when throwing a stone, the boat will begin to move towards the movement of the stone at a speed Vl relative to the ground.
Let’s write down the law of conservation of momentum: mk * Vk = ml * Vl.
Vl = mk * Vk / ml.
At the same time t, the stone must move along the boat and reach karma.
t = L / (Vk + Vl).
t = L / (Vk + mk * Vk / ml).
t = L / Vk (1 + mk / ml).
Vk = L / t * (1 + mk / ml).
Vk = 5 m / 0.5 s * (1 + 4 kg / 246 kg) = 9.84 m / s = 984 cm / s.
Answer: the speed of the stone relative to the ground should be Vк = 984 cm / s.