Answer:
58.6 m/s²
Explanation:
Applying,
s = ut+gt²/2............ equation 1
Where s = height of the rock, u = initial velocity, t = time, g = accelration due to gravity of the planet.
From the question,
Given: s = 1.55 m, t = 0.23 s, u = 0 m/s (dropped from an height)
Substitute these values into equation 1
1.55 = (0×0.23)+g(0.23²)/2
1.55 = 0.0529g/2
0.0529g = 3.1
g = 3.1/0.0529
g = 58.6 m/s²
