Newton's second law allows calculating the response for the person's acceleration while leaving the trampoline is:
-4.8 m / s²
Newton's second law says that the net force is proportional to the product of the mass and the acceleration of the body
F = m a
Where the bold letters indicate vectors, F is the force, m the masses and the acceleration
The free body diagram is a diagram of the forces without the details of the body, in the attached we can see the free body diagram for this system
[tex]F_t[/tex] -W = m a
Whera [tex]F_t[/tex] is the trampoline force
Body weight is
W = mg
We substitute
[tex]F_t[/tex] - mg = ma
a =[tex]\frac{F_t - m g}{m}[/tex]
Let's calculate
a = [tex]\frac{375 - 75 \ 9.8 }{75}[/tex]
a = -4.8 m / s²
The negative sign indicates that the acceleration is directed downward.
In conclusion using Newton's second law we can calculate the acceleration of the person while leaving the trampoline is
-4.8 m / s²
Learn more here: brainly.com/question/19860811
