The function for the height of an object dropped from a 100-meter-tall platform at time t seconds is approximated by s(t)=5t² + 100. What is the height of the object after 2 seconds?

The function given for the height \( s(t) \) of an object dropped from a 100-meter-tall platform at time \( t \) seconds is:

\[
s(t) = 5t^2 + 100
\]

To find the height of the object after 2 seconds, substitute \( t = 2 \) into the function:

\[
s(2) = 5(2)^2 + 100
\]

Calculate \( 2^2 \):

\[
2^2 = 4
\]

Now, substitute:

\[
s(2) = 5(4) + 100 = 20 + 100 = 120
\]

So, the height of the object after 2 seconds is **120 meters**.

Would you like more details or have any questions? 

Here are some related questions:

1. What is the initial height of the object?
2. How does the height change over time based on this function?
3. What is the height of the object after 5 seconds?
4. How would you find the time at which the object hits the ground?
5. Can this function be used to calculate the time of impact? Why or why not?
6. What would be the height if the object were dropped from a different height?
7. How would the height function change if gravity were different?
8. How does this height function compare to the standard height function under gravity?

**Tip:** Remember that in physics problems involving free-fall, the equation often involves the acceleration due to gravity. Double-check the constants in the function to see if they align with the scenario described.

Learn how to calculate the height of an object dropped from a 100-meter-tall platform after 2 seconds using the quadratic function. Follow a step-by-step solution with the function s(t) = 5t² + 100. Suitable for high school students studying quadratic equations and function evaluation.

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