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For a movie scene, an 85.0 kg stunt double falls 12.0 m from a building onto a large inflated landing pad. After touching the landing pad surface, it takes her 0.468 s to come to a stop. What is the magnitude of the average net force on her as the landing pad stops her?

Respuesta :

Answer:

2785.43N

Explanation:

According to Newton's second law;

[tex]F=\frac{m(v-v_1)}{t_1}..............(1)[/tex]

Given;

m = 85kg, [tex]t_1=0.468s[/tex]

v is the velocity with which she lands on the pad while [tex]v_1[/tex] is her final velocity as she comes to rest. Hence [tex]v_1=0m/s[/tex]

We obtain v from first equation of motion under gravity as follows;

[tex]v=u+gt..........(2)[/tex]

u = 0 and  [tex]g=9.8m/s^2[/tex]

hence;

v = 9.8t...........(3)

t is the time taken for her to fall through the height h =  12m, so we use the second equation of motion to find it as follows;

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