Open study

is now brainly

With Brainly you can:

  • Get homework help from millions of students and moderators
  • Learn how to solve problems with step-by-step explanations
  • Share your knowledge and earn points by helping other students
  • Learn anywhere, anytime with the Brainly app!

A community for students.

A rock dropped from a cliff covers one-third of its total distance to the ground in the last second of its fall. Air resistance is negligible. How high is the cliff?

Physics
See more answers at brainly.com
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.

Join Brainly to access

this expert answer

SIGN UP FOR FREE
distance covered inlast sec. is v+1/2g=v+5=1/3h; v=([2h/g]^1/2 -1)g; v^2=(h/5+1-2[h/5]^1/2)g^2; solve these and eliminatte v you get h
@ashwinjohn3 what would be the answer in your opinion?
ur answer will b 148.48 m

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

can i explain my work
ok use the equation....... Diplacement @ nth second=u+ a(2n-1)/2
no
i wanna use the equation\[h=ut+ \frac{ 1 }{ 2 } g t ^{2}\]
for full height u=0
and my equation become \[h=0+\frac{ 1 }{ 2 } g t ^{2}\]
\[h=\frac{ 1 }{ 2 } \times 10 \times t ^{2}=5 \times t ^{2}\]
its my 1st equation
now i wanna write another equation for 1st 2/3 h height
\[\frac{ 2 }{ 3 }h=ut+\frac{ 1 }{ 2 } \times g t ^{2}=\frac{ 1 }{ 2 } \times 10 \times (t-1)^{2}\]
now u hv to solve these 2 equation to get time t and then u will get the height h
You will get two values for t, but one of them will be less than 1 second, making it invalid, as you can see from the question "covers one-third of its total distance to the ground in the *****last second***** of its fall" Solve the two equations shamim gave you, by substituting the first one into the second, then solve for t. The second equation comes from the fact that, if it covers one third of the total distance in the last second, then it must cover the initial 2/3rds of the total distance in (t-1) seconds. \[h = \frac{ 1 }{ 2} g t ^{2}\] \[\frac{ 2 }{ 3} h = \frac{ 1 }{ 2} g (t - 1) ^{2}\]
Say the height of the cliff is H, time take for fall is T, and g is the acceleration of freefall. \[H=\frac{ 1 }{ 2 }gT ^{2}\] --- (1) \[\frac{ 2 }{3} H=\frac{ 1 }{ 2 }g(T -1)^{2}\] --- (2) Solving you will get 2 answer for T: 0.551 s and 5.45 s But T>1, Hence H = 146 m

Not the answer you are looking for?

Search for more explanations.

Ask your own question