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.

Question

anonymous · Answer

If the acceleration was constant it s 0 .

anonymous · Answer

@E.ali  No. Zero is not part of the answers.

anonymous · Answer

Well ....
Then you cant ask the question good ...
Can you ask better ?

fifciol · Answer

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 · Answer

about 1962 m/s^2

anonymous · Answer

@Fifciol do you mean19.62 m upward?

fifciol · Answer

convert cm to m you'll get then 1962 m/s^2  pointing upward

anonymous · Answer

@Fifciol  Oh, I see. Thanks.

fifciol · Answer

yw

anonymous · Answer

Ok !
I didnt know your question!