The object strikes 40.0 meters far from the base of the cliff on the ground.
Let us assume that the object strikes the ground after a time equal to 't' seconds. And horizontal distance covered by it in 't' seconds is x meters. The vertical distance traveled by it in 't' second would be equal to the height of the cliff.
Vertical distance traveled = s = 122.5m
Initial vertical speed = u = 0 m/s
Acceleration due to gravity = g = 9.8 m/s²
By using the second equation of motion:
s = ut + 1/2gt²
Put in the value
122.5 = (0)t + 1/2(9.8)t²
122.5 = 0 + 4.9t²
t² = 122.5/4.9
t² = 25
t =√25
t = 5 seconds
Horizontal speed = v = 8.0 m/s
Time taken = t = 5 s
Acceleration = 0 m/s
Using the formula, distance = speed × time
x = v × t
x = 8.0 × 5
x = 40.0 m
Therefore the object strikes 40.0 meters far from the base of the cliff on the ground.
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