# Thread: Height of a cliff

1. ## Height of a cliff

Hello All! I need help with this problem...

If a rock is is dropped from a cliff into an ocean, it travels approximately $\displaystyle 16t^2$ feet in $\displaystyle t$ seconds. If the splash is heard 4 seconds later and the speed of sound is 1100 ft./second, approximate the height of the cliff.

I thought the answer would just be: 4400 ft. I got a feeling I'm missing something. Any help would be great!

2. Originally Posted by Jonboy
Hello All! I need help with this problem...

If a rock is is dropped from a cliff into an ocean, it travels approximately $\displaystyle 16t^2$ feet in $\displaystyle t$ seconds. If the splash is heard 4 seconds later and the speed of sound is 1100 ft./second, approximate the height of the cliff.

I thought the answer would just be: 4400 ft. I got a feeling I'm missing something. Any help would be great!
HINT:
Let $\displaystyle s$ be the distance. Then the time until the sound it heard is equal to the time until the rock hits the water plus the amount of time it takes sound to travel. Use this to set up an equation.

3. Wow ThePerfectHacker that was a extremely expeditious reply!

So is this correct?

$\displaystyle s\,=\,16(4)^2\,+\,4,400$ ?

4. Originally Posted by Jonboy
Hello All! I need help with this problem...

If a rock is is dropped from a cliff into an ocean, it travels approximately $\displaystyle 16t^2$ feet in $\displaystyle t$ seconds. If the splash is heard 4 seconds later and the speed of sound is 1100 ft./second, approximate the height of the cliff.

I thought the answer would just be: 4400 ft. I got a feeling I'm missing something. Any help would be great!
Originally Posted by ThePerfectHacker
HINT:
Let $\displaystyle s$ be the distance. Then the time until the sound it heard is equal to the time until the rock hits the water plus the amount of time it takes sound to travel. Use this to set up an equation.
Originally Posted by Jonboy
Wow ThePerfectHacker that was a extremely expeditious reply!

So is this correct?

$\displaystyle s\,=\,16(4)^2\,+\,4,400$ ?
No.

It takes the sound 4 seconds to be heard, so let $\displaystyle t$ be the time of fall and $\displaystyle T$ the amount of time it takes for the sound to travel up the cliff. Let the height of the cliff above the water be h.

Then
$\displaystyle t + T = 4~s$

$\displaystyle h = 16t^2$

$\displaystyle h = (1100)T$

Solve this system for h.

-Dan

5. I have something else.

The amount of time it take for the rock to hit the water is $\displaystyle \frac{\sqrt{s}}{4}$, now the amount of time for the sound to travel up is $\displaystyle \frac{s}{1100}$.

So,
$\displaystyle \frac{\sqrt{s}}{4}+\frac{s}{1100}=4$

6. Thanks for the replies. How did you get the following TPH:

The amount of time it take for the rock to hit the water is , now the amount of time for the sound to travel up is .
?

7. Originally Posted by Jonboy
Thanks for the replies. How did you get the following TPH:
You should spend more time thinking on these:

$\displaystyle 16t^2 = s$ solve for $\displaystyle t$.

8. Originally Posted by Jonboy
Thanks for the replies. How did you get the following TPH:

?
Originally Posted by topsquark
No.

It takes the sound 4 seconds to be heard, so let $\displaystyle t$ be the time of fall and $\displaystyle T$ the amount of time it takes for the sound to travel up the cliff. Let the height of the cliff above the water be h.

Then
$\displaystyle t + T = 4~s$

$\displaystyle h = 16t^2$

$\displaystyle h = (1100)T$

Solve this system for h.

-Dan
Question answered. The only difference is I'm using h where you used s.

-Dan

9. Thank you everyone I understand now!

10. Originally Posted by Jonboy
Hello All! I need help with this problem...

If a rock is is dropped from a cliff into an ocean, it travels approximately $\displaystyle 16t^2$ feet in $\displaystyle t$ seconds. If the splash is heard 4 seconds later and the speed of sound is 1100 ft./second, approximate the height of the cliff.

I thought the answer would just be: 4400 ft. I got a feeling I'm missing something. Any help would be great!
According to this Q.,we will get 2 eqns.
t1=x/16t...........1 , where t1(time taken by the stone to reach sea),16t(Speed of the stone)

t2=x/1100............2 , where t2(time taken by the sound to reach cliff)

from 1 and 2,we get t1+t2=x/16t+x/1100......3 and given t1+t2=4 seconds.

Therefore, eqn 3 changes to
4=x/16t+x/1100...........4 , which is a 2 variable eqn. so cannot be solved without a extra eqn. or we have to solve it by Hit and Trial or this Q. might have options to solve.
But it is sure x cannot be equal to 4400.(See the eqn. 4)
Taking t=1
, and solving eqn 4, we will get,x=63.08 ft (approx)
Taking t=2, and solving eqn 4, we will get,x=124.38ft (approx)
Taking t=3, and solving eqn 4, we will get,x=183.97 ft (approx)

choose ur ans.

regards
upendra

11. Originally Posted by Upendra Bhagat
According to this Q.,we will get 2 eqns.
t1=x/16t...........1 , where t1(time taken by the stone to reach sea),16t(Speed of the stone)
No because the stone will accelerate downward. This equation assumes the speed of the stone is constant, so it is not physically valid.

-Dan