A girl jumps off the mall as part of a stunt for shipping week festival. She falls 30.0m before she hits the big airbag. She squishes the airbag by 2.30m.
a) With what speed speed did she hit the air bag?
b) What acceleration did she feel while she was squishing the airbag ? this must be treated as a 2nd trip!!!
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a) When the girl jumps off the mall, her initial velocity is zero. Before she hits the airbag, her acceleration is only due to gravity, so her final speed `v_f` (the speed just before she hits the airbag) can be found from
`v_f ^2 - v_i ^2 = 2gh` , where h is the distance which she falls before hitting the airbag. Since the initial velocity `v_i = 0` , then
`v_f ^2 = 2gh` and
`v_f = sqrt(2gh)` .
Plugging in g = 9.8 m/s^2 and h = 30 meters, we get
`v_f = sqrt(2*9.8*30) = 24.23 m/s`
The speed with which the girl hits the airbag is 24.33 m/s.
b) After the girl hits the airbag, there are two forces acting on her: gravity and the force from the airbag, so her acceleration is unknown. Since she squished the airbag by 2.3 m before coming to a complete stop, this is the distance she traveled with this acceleration. The same equation as above can be used to describe this motion, if we assume that the acceleration is constant:
`v_f ^2 - v_i ^2 = 2ad` .
Now, the final velocity is `v_f = 0` , the initial velocity is the velocity with which the girl hit the bag, found in a): `v_i = 24.23 m/s` , and the distance is d = 2.3 meters. Plugging in these values, we can find the acceleration:
`0^2 - 588 = 2*a*2.3`
`a = -588/(2*2.3) = -127.82 m/s^2`
The negative sign indicates that the acceleration is directed upward, opposite the girl's velocity, which is consistent with the fact that she has slowed down to a stop after hitting the bag. Notice how much greater this acceleration is than the acceleration due to gravity, 9.8 m/s^2 !
The girl felt the acceleration of 127.82 m/s^2 while squishing the airbag.
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