anonymous
  • anonymous
A student drops a rock from a bridge to the water 12 m below. With what speed does the rock strike the water?
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.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
raffle_snaffle
  • raffle_snaffle
What are you givens?
raffle_snaffle
  • raffle_snaffle
|dw:1378438909942:dw|
raffle_snaffle
  • raffle_snaffle
Are you given a time?

Looking for something else?

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

More answers

raffle_snaffle
  • raffle_snaffle
or a mass of the stone?
anonymous
  • anonymous
No that's the whole problem
raffle_snaffle
  • raffle_snaffle
You need one more piece of info in order to figure out the speed of the rock when it hits the water.
anonymous
  • anonymous
Ok lets see the data given, distance=S=12 metres and a=g=9.8\[m/s ^{2}\] initial velocity =vi=0 and final velocity =vf=??, according to newton equation of motion \[vf ^{2}-vi ^{2}=2aS\] so putting values we get \[vf ^{2}-(0)^{2}=2\times9.8\times12\] \[vf ^{2}=235.2\] taking square root both side \[\sqrt{vf ^{2}}=\sqrt{235.2}\] \[vf=15.336 metres/second\] so the final velocity will be 15.3 m/s
goformit100
  • goformit100
"Welcome to OpenStudy. I can guide regarding this useful site; ask your doubts from me, for it you can message me. Please use the chat for off topic questions. And remember to give the helper a medal, by clicking on "Best Answer". We follow a code of conduct, ( http://openstudy.com/code-of-conduct ). Please take a moment to read it."

Looking for something else?

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