## Robinronaldo Group Title 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? one year ago one year ago

• This Question is Open
1. exploringphysics Group Title

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 Group Title

3. shamim Group Title

ur answer will b 148.48 m

4. shamim Group Title

can i explain my work

5. ashwinjohn3 Group Title

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

6. shamim Group Title

no

7. shamim Group Title

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

8. shamim Group Title

for full height u=0

9. shamim Group Title

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

10. shamim Group Title

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

11. shamim Group Title

its my 1st equation

12. shamim Group Title

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

13. shamim Group Title

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

14. shamim Group Title

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

15. agent0smith Group Title

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 Group Title

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