The function provided, $s(t) = 5t + 100$, describes the height of the object at time $t$ seconds after being dropped.

To find the height after 2 seconds, substitute $t = 2$ into the function:

$s(2) = 5(2) + 100 = 10 + 100 = 110 \text{ meters}$

However, this result does not match any of the provided choices. This discrepancy suggests that the function might still be incorrect or that the choices provided don't align with the function you intended. The typical function for an object dropped from a height under gravity would usually involve $t^2$.

Could you please confirm or provide more context?

Here are 8 related questions:

1. Could there be a typo in the function $s(t) = 5f + 100$?
2. What would the height be after 3 seconds using the same function?
3. What is the physical meaning of the coefficient 5 in the equation?
4. How would the function change if the object was thrown upwards?
5. Is air resistance considered in this function?
6. How would the height be calculated if the initial height was different?
7. What is the general form of the equation of motion under gravity?
8. How would the function change if the object were dropped on a different planet?

Tip: Always double-check the function for any possible typos or misinterpretations, especially in physics-related problems.