12. A rock on the surface of Mars is projected vertically upwards with an intial speed of 9.4ms The
rock rises to a height of 12 m above the surface.
Assume there is no atmosphere on Mars
What is the acceleration of free fall near the surface of Mars?

The  acceleration of free fall near the surface of Mars is .

Given data:

The magnitude of initial velocity of rock is, u = 9.4 m/s.

The height raised up by rock is, h = 12 m.

Here, we need to find the acceleration of free fall, which is nothing but the gravitational acceleration. So we will use the Third kinematic equation of motion, which is given as,

Solving as,

Thus, we can conclude that the  acceleration of free fall near the surface of Mars is .

Learn more about the gravitational acceleration here:

link

3.68 m/s².

Explanation:

From the question given above, the following data were obtained:

Initial velocity (u) = 9.4 ms¯¹

Maximum height (h) reached = 12 m

Final velocity (v) = 0 (at maximum height)

Acceleration of free fall (g) =?

v² = u² – 2gh (since the rock is going against gravity)

0 = 9.4² – 2 × g × 12

0 = 88.36 – 24g

Collect like terms

0 – 88.36 = –24g

–88.36 = –24g

Divide both side by –24

g = –88.36 / –24

g = 3.68 m/s²

Thus, the acceleration of free fall near the surface of Mar is 3.68 m/s².

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