A projectile weighing 3 kg, fired at an angle of 45 degrees to the horizon, flew a horizontal distance of 10 km

A projectile weighing 3 kg, fired at an angle of 45 degrees to the horizon, flew a horizontal distance of 10 km. What will be the kinetic energy of the projectile just before it hits the Earth? Neglect air resistance.

m = 3 kg.

∠α = 45 “.

g = 9.8 m / s ^ 2.

S = 10 km = 10000 m.

Ek -?

The kinetic energy of the body Ek is determined by the formula: Ek = m * V ^ 2/2, where m is the mass of the body, V is the velocity at the moment of the fall of the projectile.

Since friction is neglected, according to the law of conservation of energy, the speed of the projectile at the moment of falling is equal to the speed of the projectile at the moment of the shot.

The projectile moves horizontally evenly, the flight range S will be determined by the formula: S = V * cosα * t.

The projectile moves vertically with free fall acceleration g, the flight time t will be determined by the formula: t = 2 * V * sinα / g.

S = V * cosα * 2 * V * sinα / g = V ^ 2 * 2 * cosα * sinα / g = V ^ 2 * sin2 * α / g.

V ^ 2 = S * g / sin2.

Ek = m * S * g / sin2 * 2.

Ek = 3 kg * 10000 m * 9.8 m / s ^ 2 / sin90 “* 2 = 147000 J.

Answer: the kinetic energy of the projectile at the moment of impact on the ground Ek = 147,000 J.