Answer:
10.6cm
Explanation:
We are given 5.3cm below the starting point (spring extension).
Therefore, to find static vertical equilibrium, we use the equation:
kx = mg
Where:
k = spring constant =
=mg/5.3 kg/s²
We are told the object was dropped from rest.
Therefore:
loss in potential energy = gain in spring p.e
Let's use the expression:
mgx = ½kx²
We are asked to find the stretch at maximum elongation x.
To find x, we make x subject of the formula.
Therefore, we have:
x = 2mg/k (after rearranging the equation above)
x = (2mg) / (mg/5.3)
x = 10.6cm
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In this problem, we have an <u>initial velocity</u> 
jump upwards (opposite direction to gravity) with the <u>acceleration</u> due to gravity 
, finally the player will reach to <u>final velocity</u> in the air 
Using the equation of motion, the third equation states that:
where h is the height , the distance which the player reach in air
Thus
