A rock, dropped from rest near the surface of an atmosphere-free planet, attains a speed of 20.0 m/s after falling 8.0 meters. What is the magnitude of the acceleration due to gravity on the surface of this planet

a = 25 m/s^2

the magnitude of the acceleration due to gravity on the surface of this planet is 25 m/s^2

Explanation:

Applying the equation of motion;

v^2 = u^2 + 2as 1

Where;

v = final speed = 20 m/s

u = initial speed = 0 (released from rest)

a = acceleration due to gravity on the planet

s = distance covered = 8.0 m

Since u = 0, equation 1 becomes;

v^2 = 2as

Making a the subject of formula;

a = (v^2)/2s

Substituting the values;

a = (20^2)/(2×8)

a = 25 m/s^2

the magnitude of the acceleration due to gravity on the surface of this planet is 25 m/s^2

divide the right side by the left side

0.024÷1.2=0.02 for example