Answer:
Explanation:
Given
mass of spring 
extension in spring 
downward velocity 
Position in undamped free vibration is given by

where 
also 



it is given


substituting values we get







Potential Energy= 24m * 14kg * 9.8N/kg = 3292.8J
Answer:
The gravitational acceleration of the planet is, g = 8 m/s²
Explanation:
Given data,
The distance the object falls, s = 144 m
The time taken by the object is, t = 6 s
Using the III equations of motion
S = ut + ½ gt²
∴ g = 2S/t²
Substituting the given values,
g = 2 x 144 /6²
= 8 m/s²
Hence, the gravitational acceleration of the planet is, g = 8 m/s²
Brown black fur and medium tail
I’ve done this before the answer is B