Question

12.

Answer 1

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.