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.
