1 2 3 4 5 6 - 8 9 I 1 The acceleration due to gravity is the constant of variation. What is the acceleration due to gravity of a falling object? $$ 4.9 \frac{\mathrm{~m}}{\mathrm{~s}^{2}} $$ $$ 9.8 \frac{\mathrm{~m}}{\mathrm{~s}^{2}} $$ $$ 10.2 \frac{\mathrm{~m}}{\mathrm{~s}^{2}} $$ $$ 19.6 \frac{\mathrm{~m}}{\mathrm{~s}^{2}} $$

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UpStudy AI · Accepted Answer

The acceleration due to gravity is a well-known constant that describes the rate at which an object accelerates towards the Earth when in free fall.

The standard value for the acceleration due to gravity near the surface of the Earth is:

$$
g = 9.8 \frac{\mathrm{~m}}{\mathrm{~s}^{2}}
$$

This means that, in the absence of air resistance, an object will accelerate downwards at this rate.

Now, let's analyze the options provided:

1. $$ 4.9 \frac{\mathrm{~m}}{\mathrm{~s}^{2}} $$ - This is half of the standard value and is not correct.
2. $$ 9.8 \frac{\mathrm{~m}}{\mathrm{~s}^{2}} $$ - This is the correct value for the acceleration due to gravity.
3. $$ 10.2 \frac{\mathrm{~m}}{\mathrm{~s}^{2}} $$ - This is slightly higher than the standard value and is not correct.
4. $$ 19.6 \frac{\mathrm{~m}}{\mathrm{~s}^{2}} $$ - This is double the standard value and is not correct.

Thus, the correct answer is:

$$
\boxed{9.8 \frac{\mathrm{~m}}{\mathrm{~s}^{2}}}
$$
The acceleration due to gravity is $$ 9.8 \frac{\mathrm{~m}}{\mathrm{~s}^{2}} $$.

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