The projectile flew from a long-range cannon with an initial speed of 1000m / s at an angle of 30 degrees

The projectile flew from a long-range cannon with an initial speed of 1000m / s at an angle of 30 degrees to the horizon. How long was the projectile in the air? How far from the bullet will it fall to the ground? The cannon and the drop point are on the same horizontal. What speed will the projectile have when it falls to the ground?

V0 = 1000 m / s.

∠α = 30 °.

g = 10 m / s2.

L -?

V -?

The movement of the projectile can be divided into two types: horizontally, it moves uniformly at a speed Vx = V0 * cosα, vertically uniformly accelerated with the acceleration of gravity g.

The flight range L is expressed by the formula: L = Vх * t = V0 * cosα * t, where V0 is the projectile departure speed, t is the projectile flight time.

g = V0 * sinα / t1, where t1 is the time for the grenade to rise to the maximum height.

t1 = V0 * sinα / g.

t = 2 * t1 = 2 * V0 * sinα / g.

L = V0 * cosα * 2 * V0 * sinα / g = V02 * sin2α / g.

L = (1000 m / s) 2 * sin2 * 30 ° / 10 m / s2 = 86600 m.

If the force of air friction is neglected, then, according to the law of conservation of total mechanical energy, the velocity at the moment of projectile departure is equal to the velocity of the projectile upon landing.

V = V0 = 1000 m / s.

Answer: the projectile flew a distance L = 86600 m, at the time of landing it had a speed of V = 1000 m / s.