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The problem states:

"A free-falling object drops from a height of 50 meters, so its height at time \( t \) is given by \( h(t) = 50 - 4.9t^2 \). What is its velocity at \( t = 2 \)?"

### Solution:

To find the velocity of the object at \( t = 2 \), we need to determine the derivative of the height function \( h(t) \) with respect to time \( t \). The velocity \( v(t) \) is the first derivative of \( h(t) \):

\[
h(t) = 50 - 4.9t^2
\]

1. **Differentiate \( h(t) \) to find \( v(t) \):**

   \[
   v(t) = \frac{dh}{dt} = -2 \times 4.9 \times t = -9.8t
   \]

2. **Substitute \( t = 2 \) into \( v(t) \):**

   \[
   v(2) = -9.8 \times 2 = -19.6 \, \text{m/s}
   \]

### Answer:
The velocity of the object at \( t = 2 \) seconds is \( -19.6 \, \text{m/s} \), where the negative sign indicates the object is moving downward.

Would you like further explanation or have additional questions?

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Here are some related questions for deeper understanding:
1. What would be the velocity of the object at \( t = 3 \) seconds?
2. How does the acceleration of the object relate to the function \( h(t) \)?
3. What would be the height of the object at \( t = 2 \) seconds?
4. How would the function change if the object was thrown upwards initially?
5. What assumptions are made in this free-fall calculation?

**Tip:** In physics, the derivative of position with respect to time gives velocity, and the derivative of velocity with respect to time gives acceleration.

Sebuah benda jatuh bebas dari ketinggian 50 m, sehingga tingginya pada saat t adalah h(t) = 50 - 4.9t^2. Berapakah kecepatannya pada saat t = 2?

Learn how to calculate the velocity of a free-falling object from a height of 50 meters, with height given by h(t) = 50 - 4.9t^2, at t = 2 seconds. This problem involves key calculus and physics concepts suitable for high school students.

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