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
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The average velocity of the object in moving from point-B to point-C is
(the straight-line distance and direction from point-B to point-C)
divided by
(the time the object takes to make the trip) .
The energy of the baby is gravitational potential energy, and it is equal to the weight of the baby times its height from the ground:
where
mg=20 N is the weight (the mass times the gravitational acceleration)
h=1.5 m is the height from the ground
If we plug the numbers into the equation, we find
Explanation:
hope this helps you out if not im sorry
