Question

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?

Answer 1

Answer:

2785.43N

Explanation:

According to Newton's second law;

$F=\frac{m(v-v_1)}{t_1}..............(1)$

Given;

m = 85kg, $t_1=0.468s$

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

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

$v=u+gt..........(2)$

u = 0 and  $g=9.8m/s^2$

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;