The answer to your question is TWICE AS GREAT
The distance a dropped object falls, with gravity and no air resistance:
Distance = (1/2) (acceleration) (falling time)²
Without air resistance, the horizontal motion has no effect on the fall.
Acceleration of Earth gravity = 9.8 m/s²
Distance = (1/2) (acceleration) (falling time)²
Distance = (1/2) (9.81 m/s²) (3.0 s)²
Distance = (0.5) x (9.81 m/s²) x (9.0 s²)
Distance = (0.5 x 9.81 x 9.0) (m-s² / s²)
Distance = 44.15 meters
We don't care how fast the bird was flying horizontally. It doesn't change anything. (It DOES determine how far ahead of the drop point the clam hits the ground. Most problems like this ask for that distance. This one didn't.)
Answer:
is changing in direction, but constant in magnitude
Explanation:
This question is a bit tricky since the velocity of the satellite is changing, but the speed is constant.
Speed is simply a measure of how fast you are going. It doesn't matter where you're going, just how quickly.
Velocity, on the other hand, does care about which direction you're going. For example, it could be then when you travel right, your velocity is positive, and when you travel left, your velocity is negative. This is the similar for a 2D shape like a circular orbit
Since we know velocity is changing, there must be acceleration which changes that velocity (since acceleration <em>is</em><em> </em>the change in velocity: going from 0 to 60 mph, for example)
Thus, with a non-zero net acceleration, we know that there must be a force that is changing in direction, but constant in magnitude (since the orbit is a circle, and always attracted to the center of the Earth at equal distance).
