anonymous
  • anonymous
A rock falls off of a 2.00 m high wall. When it hits the ground, the ground compresses a distance of 1.00 cm as it comes to a stop. What was the acceleration of the rock as it came to a stop after contacting the ground? Assume the motion during the stop is with constant acceleration.
Physics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
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.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
If the acceleration was constant it s 0 .
anonymous
  • anonymous
@E.ali No. Zero is not part of the answers.
anonymous
  • anonymous
Well .... Then you cant ask the question good ... Can you ask better ?

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

Fifciol
  • Fifciol
Since a is constant \[a=\frac{ v }{ t }\Rightarrow a^2=\frac{v^2}{t^2}\] \[d=\frac{ at^2 }{2 } \Rightarrow t^2=\frac{2d}{a}\] Conservation of mechanical energy \[mgh=\frac{ mv^2 }{ 2 } \Rightarrow v^2=2gh\]so \[a^2=\frac{ 2gha }{ 2d} \Rightarrow a=\frac{ gh }{ d }\]
Fifciol
  • Fifciol
about 1962 m/s^2
anonymous
  • anonymous
@Fifciol do you mean19.62 m upward?
Fifciol
  • Fifciol
convert cm to m you'll get then 1962 m/s^2 pointing upward
anonymous
  • anonymous
@Fifciol Oh, I see. Thanks.
Fifciol
  • Fifciol
yw
anonymous
  • anonymous
Ok ! I didnt know your question!

Looking for something else?

Not the answer you are looking for? Search for more explanations.