Answer:
a = 1.1 10⁵ m / s²
Explanation:
This is a momentum exercise, where we use the relationship between momentum and momentum
I = ∫ F dt = Δp
= p_f - p₀
as they indicate that the ball bounces at the same height, we can assume that the moment when it reaches the ground is equal to the moment when it bounces, but in the opposite direction
F t = 2 (m v)
therefore the average force is
F = 2 m v / t
where in general the mass of the ball unknown, the velocity of the ball can be calculated using the conservation of energy
starting point. Done the ball is released with zero initial velocity
Em₀ = U = mgh
final point. Upon reaching the ground, just before the deformation begins
Em_f = K = ½ m v²
energy is conserved in this system
Em₀ = Em_f
m g h = ½ m v²
v = √ (2gh)
This is the velocity of the body when it reaches the ground, so the force remains
F = 2m √(2gh) /t
where the height of the person's chest is known and the time that the impact with the floor lasts must be estimated in general is of the order of milli seconds
knowing this force let's use Newton's second law
F = m a
a = F / m
a = 2 √(2gh) / t
We can estimate the order of magnitude of this acceleration, assuming the person's chest height of h = 1.5 m and a collision time of t = 1 10⁻³ s
a = 2 √ (2 9.8 1.5) / 10⁻³
a = 1.1 10⁵ m / s²
