Answer:
The time taken by the brick to hit the ground, t = 0.84 s
Explanation:
Given that,
A brick falls from a height, h = 3.42 m
The initial velocity of the brick is zero.
Since the brick is under free-falling. The time equation of a free-falling body when the displacement is given is
t = 
where,
h - height from surface in meters
g - acceleration due to gravity
on substituting the values in the above equation,
t = 
= 0.84 s
Hence, time taken by the brick to hit the ground is t = 0.84 s
Answer:
The maximum height above the point of release is 11.653 m.
Explanation:
Given that,
Mass of block = 0.221 kg
Spring constant k = 5365 N/m
Distance x = 0.097 m
We need to calculate the height
Using stored energy in spring
...(I)
Using gravitational potential energy
....(II)
Using energy of conservation




Where, k = spring constant
m = mass of the block
x = distance
g = acceleration due to gravity
Put the value in the equation


Hence, The maximum height above the point of release is 11.653 m.