https://r2.erweima.ai/mathbot/image/1725643901214ra62v790.jpg

The equation in the image appears to be:

\[
0.0162 = 0.0943^2 \times 0.0960 \times \left[ \frac{t^2 \times 9.80665}{a \times 1.5000} - 1 \right]
\]

Let's break down each part of the equation:

1. **Left-hand side (LHS)**: \( 0.0162 \), a constant.
2. **Right-hand side (RHS)**:
    - \( 0.0943^2 \), the square of \( 0.0943 \).
    - \( 0.0960 \), multiplied directly after \( 0.0943^2 \).
    - A fraction within square brackets: \( \frac{t^2 \times 9.80665}{a \times 1.5000} - 1 \), where:
        - \( t^2 \) represents time squared.
        - \( 9.80665 \) represents the acceleration due to gravity (m/s²).
        - \( a \times 1.5000 \), with \( a \) representing an unknown constant or variable.

I can now help you solve for \( t \), \( a \), or simplify it further if you provide more context.

Would you like to proceed with solving for a particular variable, or need more explanation? 

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### Related Questions:
1. How do we isolate \( t \) in this type of equation?
2. How does gravity, \( 9.80665 \), affect calculations involving time and acceleration?
3. What does squaring a term (like \( t^2 \)) imply in physical scenarios?
4. What units might be associated with each constant or variable?
5. How can we solve this equation using numerical methods?

### Tip:
Always verify if your variables and constants are in compatible units when dealing with physical equations!

0.0162 = 0.0943^2 × 0.0960 × [ (t^2 × 9.80665) / (a × 1.5000) - 1 ]

Learn how to solve an equation involving gravity (9.80665 m/s²), time squared, and algebraic manipulation. This problem is suitable for students in Grades 10-12 and covers key concepts in both algebra and physics.

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