To answer this question, we must first understand the scenario. In this case, a rock is being thrown horizontally off a bridge, which means it is moving only along the horizontal axis with an initial velocity of 3.5 m/s.
We need to determine the magnitude of its velocity as soon as possible (ASAP), which is immediately after it was thrown.
Here's how to approach it:
1. As we're looking at the situation right after the rock is thrown, the only active velocity component is the horizontal one. Why? Because there's not been enough time for the acceleration due to gravity to significantly act on it. Therefore, its vertical velocity component is at that moment still essentially zero.
2. Meanwhile, the rock's horizontal velocity remains constant at 3.5 m/s. This is because there are no forces acting horizontally to either slow it down or speed it up. This means that its horizontal velocity is the same as its initial velocity of 3.5 m/s.
3. The velocity of an object can be thought of as a vector with horizontal and vertical components. Right after the rock is thrown horizontally, its vertical velocity is zero and its horizontal velocity is 3.5 m/s.
4. Remember, the magnitude of a vector is calculated as the square root of the sum of the squares of its components. However, since the vertical velocity is zero, the magnitude of the velocity is just the absolute value of the horizontal component.
5. We thus find that the magnitude of the velocity of the rock right after it was thrown (ASAP) is 3.5 m/s.
So, the magnitude of the rock's velocity as soon as it leaves the thrower's hand is 3.5 m/s.