Referring to the above two questions, the acceleration due to gravity is given by

10 m / s e c^{2}

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5 m / s e c^{2}

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20 m / s e c^{2}

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2.5 m / s e c^{2}

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Solution

The correct option is A $10 m / s e c^{2}$
$a_{x} = \frac{d}{d t} (v_{x}) = 0, a_{y} = \frac{d}{d t} (v_{y}) = - 10 m / s^{2}$
$∴ N e t a c c e l e r a t i o n a = \sqrt{a_{x}^{2} + a_{y}^{2}} = \sqrt{0^{2} + 10^{2}} = 10 m / s^{2}$

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Similar questions

0.2 m / s^{2} .

Calculate the value of acceleration due to gravity at a point 5.0 km above earth's surface.

9.8 m s^{- 2}

4.9 m s^{- 2}

= 6.4 \times 10^{6} m

9.8 {ms}^{- 2}

4.9 {ms}^{- 2}

= 6.4 \times 10^{6} m

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