Linear Motion: calculating height of cliff given speed of sound and time

[rahrahrah1]

Homework Statement

A student standing on top of a cliff drops a rock down below into the water and hears it splash 3 seconds later. The speed of sound is 330m/s, what is the height of the cliff

Homework Equations

v= d/t
v2 = v1 + at
d = v2-1/2 at2
d = (v1 + v2)t
v2 squared = v1 squared + 2ad
d = v1t + at2

The Attempt at a Solution

when I first saw this, I thought of echos
v=d/t
330 = d/3
d=990
and then divide by two since echo
d= 990/2
d= 495
however, that is horizontal distance not vertical, so I listed my knowns but am unsure as what to do with the 330 m/s

knowns for rock
v1= 0 (since he dropped the rock)
v2=
a = -9.8 m/s2
t = 3s
d =

knowns for speed of sounds
v= 330 m/s
t = 3s
d= ?

then setting the distance of both of these equal to each other and solving for d?

 

[PeterO]

Homework Statement

A student standing on top of a cliff drops a rock down below into the water and hears it splash 3 seconds later. The speed of sound is 330m/s, what is the height of the cliff

Homework Equations

v= d/t
v2 = v1 + at
d = v2-1/2 at2
d = (v1 + v2)t
v2 squared = v1 squared + 2ad
d = v1t + at2

The Attempt at a Solution

when I first saw this, I thought of echos
v=d/t
330 = d/3
d=990
and then divide by two since echo
d= 990/2
d= 495
however, that is horizontal distance not vertical, so I listed my knowns but am unsure as what to do with the 330 m/s

knowns for rock
v1= 0 (since he dropped the rock)
v2=
a = -9.8 m/s2
t = 3s
d =

knowns for speed of sounds
v= 330 m/s
t = 3s
d= ?

then setting the distance of both of these equal to each other and solving for d?

Some of the time [most?] is taken up with the rock falling. A small amount at the end is the sound traveling up at 330 m/s.

For example, if you drop something and it falls for 3 seconds, it falls a little less than 45m

At 330 m/s, the sound would take about 0.15 seconds to come back the 45 m

We thus know the cliff is less than 45 m high.

 

[rahrahrah1]

hmm, so if I change my variables in the "knowns for rock"
v1: 0
v2:
a: -9.8m/s
t: t-3
d:

and sub into solve for d...

d rock = v1t + 0.5at squared
= (0)t + 0.5(-9.8)(t-3) squared
= -4.9tsquared + 29.4t -44.1

dsound =vt
= (330)t

330t = -4.9squared + 29.4t -44.1
= -4.9squared -300.6t - 44.1
= [-b +/- √(b squared-4ac)]2a
= [300.6 +/- √(89496)]-9.8
t= -61.2 or -.15
I get the same 0.15 time however, mine is negative for some reason..

 

[PeterO]

hmm, so if I change my variables in the "knowns for rock"
v1: 0
v2:
a: -9.8m/s
t: t-3
d:

and sub into solve for d...

d rock = v1t + 0.5at squared
= (0)t + 0.5(-9.8)(t-3) squared
= -4.9tsquared + 29.4t -44.1

dsound =vt
= (330)t

330t = -4.9squared + 29.4t -44.1
= -4.9squared -300.6t - 44.1
= [-b +/- √(b squared-4ac)]2a
= [300.6 +/- √(89496)]-9.8
t= -61.2 or -.15
I get the same 0.15 time however, mine is negative for some reason..

I think that first time should be 3-t

That change of sign may fix things

 

[asteriatic]

I would approach this problem by first clarifying the question and making sure all the necessary information is provided. It is important to note that the speed of sound is not relevant in this scenario, as the time taken for the rock to hit the water is independent of the speed of sound. Additionally, the given equations are not directly applicable to this problem.

To calculate the height of the cliff, we can use the equation d = v1t + 1/2at^2, where d is the vertical distance, v1 is the initial velocity (in this case, it is 0 since the rock was dropped), a is the acceleration due to gravity (9.8 m/s^2), and t is the time taken for the rock to hit the water (3 seconds).

Plugging in the known values, we get:

d = (0)(3) + 1/2(9.8)(3)^2
d = 0 + 1/2(29.4)
d = 14.7 meters

Therefore, the height of the cliff is approximately 14.7 meters. It is important to note that this calculation assumes no air resistance and a constant acceleration due to gravity. In reality, the height may be slightly different due to these factors.