anonymous
  • anonymous
A 5.5 kg bowling ball initially at rest is dropped from the top of a 12 m building. It hits the ground 1.75 s later. Find the net external force on the falling ball.
Physics
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anonymous
  • anonymous
A 5.5 kg bowling ball initially at rest is dropped from the top of a 12 m building. It hits the ground 1.75 s later. Find the net external force on the falling ball.
Physics
schrodinger
  • schrodinger
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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yrelhan4
  • yrelhan4
s=1/2at^2 you have s and t find a. net force=ma. plug and solve.
anonymous
  • anonymous
What is s?
yrelhan4
  • yrelhan4
s-->distance travelled a--> acceleration due to gravity here t-->time taken

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anonymous
  • anonymous
Okay, thanks:)
yrelhan4
  • yrelhan4
cheers!
anonymous
  • anonymous
I 431.2N but that isn't an option..
yrelhan4
  • yrelhan4
43.12 N is the answer.
anonymous
  • anonymous
So this is what I did.. s=12m a=? t=1.75s 12=((1.75)^2a)/2 12=(3.0625a)/2 24=3.0625a a=7.84 netforce=55*7.84 netforce=431.2N
yrelhan4
  • yrelhan4
Net acceleration. a = 2*12/1.75² Net force ma = 5.5* 2*12/1.75² = 43.12 N
anonymous
  • anonymous
I get what your saying now. But the only problem is my choices are 120N, 54N, & 0N. But I can rule out 0N automatically
yrelhan4
  • yrelhan4
hmm. no idea. sorry.
anonymous
  • anonymous
Its okay, thanks for your help though :) I really do appreciate it
yrelhan4
  • yrelhan4
you're welcome. good luck. :)

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