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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?

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Respuesta :

Eduard22sly

Answer:

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.

snehashish65

The  acceleration of free fall near the surface of Mars is [tex]3.68 \;\rm m/s^{2}[/tex].

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,

[tex]v^{2}=u^{2}+2gh[/tex]

Solving as,

[tex]0^{2}=9.4^{2}+(2 \times -g \times 12)\\\\g = 3.68 \;\rm m/s^{2}[/tex]

Thus, we can conclude that the  acceleration of free fall near the surface of Mars is [tex]3.68 \;\rm m/s^{2}[/tex].

Learn more about the gravitational acceleration here:

https://brainly.com/question/3663429

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