A space probe is fired as a projectile from the earth's surface with an initial speed of 2.44 104m/s. what will its speed be when it is very far from the earth? ignore atmospheric friction and the rotation of the earth.

Projectile motion is characterized by an arc-shaped direction of motion. It is acted upon by two vector forces: horizontal component and vertical component. The horizontal component is in constant velocity motion, while the vertical component is in constant acceleration motion. These two motions are independent of each other.

Now, the total velocity of the space probe at the end of the projectile motion is determined through this equation:

V = √ (Vx² + Vy²)

where Vx is the velocity in the horizontal direction and Vy is the velocity in the vertical direction.

Let's find Vx first. Assuming that the space probe was launched at an angle horizontal the Earth's surface, the launching angle is 0°. Thus, the initial velocity is 2.44*10⁴ m/s.

For Vy, the free falling motion is

Vy = √ (2gh), where g is 9.81 m/s² and h is the distance traveled vertically by the space probe which represents the radius of the Earth equal to 6.37*10⁶ meters. Therefore,

V = √{ (2.44*10⁴) ² + [√ (2*9.81*6.37*10⁶) ]²}

V = 26,839.14 m/s