the function v(t) = -9,8 +5 gives the instantaneous velocity (in m/sec) of an object thrown upward with an initial velocity of 5m/sec. At what time t does the object start falling?

To determine the time \( t \) when the object starts falling, we need to find the moment when the instantaneous velocity \( v(t) \) becomes zero. This is because the object stops moving upward and starts falling downward when its velocity changes from positive to negative.

Given the function:

\[
v(t) = -9.8t + 5
\]

Set \( v(t) = 0 \) to find the time \( t \) when the velocity is zero:

\[
0 = -9.8t + 5
\]

Now, solve for \( t \):

\[
9.8t = 5
\]

\[
t = \frac{5}{9.8}
\]

\[
t \approx 0.51 \text{ seconds}
\]

So, the object starts falling at approximately \( t = 0.51 \) seconds.

Would you like more details or have any questions about this problem?

Here are 5 related questions to expand your understanding:
1. What is the significance of the sign of velocity in this context?
2. How would you find the maximum height reached by the object?
3. What would the velocity be after 1 second?
4. How long will it take for the object to hit the ground if it was thrown from a height of 10 meters?
5. How would the function change if the object were thrown downward instead?

**Tip:** Remember, the object starts falling when the velocity is zero, which is the point where it changes direction.

Discover how to find the time when an object starts falling using the equation of motion for velocity under gravity. This problem involves kinematics and instantaneous velocity, relevant for high school students.

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