## Robinronaldo 2 years ago 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?

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1. exploringphysics

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

2. Robinronaldo

@ashwinjohn3 what would be the answer in your opinion?

3. shamim

ur answer will b 148.48 m

4. shamim

can i explain my work

5. ashwinjohn3

ok use the equation....... Diplacement @ nth second=u+ a(2n-1)/2

6. shamim

no

7. shamim

i wanna use the equation$h=ut+ \frac{ 1 }{ 2 } g t ^{2}$

8. shamim

for full height u=0

9. shamim

and my equation become $h=0+\frac{ 1 }{ 2 } g t ^{2}$

10. shamim

$h=\frac{ 1 }{ 2 } \times 10 \times t ^{2}=5 \times t ^{2}$

11. shamim

its my 1st equation

12. shamim

now i wanna write another equation for 1st 2/3 h height

13. shamim

$\frac{ 2 }{ 3 }h=ut+\frac{ 1 }{ 2 } \times g t ^{2}=\frac{ 1 }{ 2 } \times 10 \times (t-1)^{2}$

14. shamim

now u hv to solve these 2 equation to get time t and then u will get the height h

15. agent0smith

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. $h = \frac{ 1 }{ 2} g t ^{2}$ $\frac{ 2 }{ 3} h = \frac{ 1 }{ 2} g (t - 1) ^{2}$

16. chongkhengwye

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