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Complex physics kinematics problems help

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Complex physics kinematics problems help!!!

I'm having difficulty with these physics problems. I would really appreciate it if someone could help me.

Homework Statement


1) Steven throws a rock off a cliff, giving it a velocity of 8.3 m/s [down). He heard the splash when the rock hit the river below, exactly 6.9s after he threw the rock. How high is the cliff above the river? Assume the speed of sound equals 330 m/s.


Homework Equations


I've been given
v = d/t
a = (v2-v1)/t
d = ((v2 + v1)/2) t
d = (v1)t + 1/2a(t^2)
d = (v2)t - 1/2a(t^2)
d = (v2^2 - v1^2)/2a

The Attempt at a Solution


I don't have the slightest clue how to begin solving this question. I don't know how the speed of sound ties into the solution
 
Last edited:

Answers and Replies

  • #2
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The time that it takes until he hears the splash (6.9 seconds) is the time is takes for the rock to travel to the bottom of the cliff plus the time it takes for the sound of the splash to travel back to the top. That is to say:

[itex]\displaystyle T_r + T_s = 6.9 seconds[/itex]
Tr is the travel time of the rock, and Ts is the travel time of the sound.

You should start by making an equation (you can use the equations you listed) for the travel time of the rock (Tr) as a function of distance. Then similarly, you can make an equation for the travel time of the sound(Ts).

Hope that helps. If you have more trouble, feel free to ask but it helps if you post your progress to give an idea of where you're stuck.
 
  • #3
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The time that it takes until he hears the splash (6.9 seconds) is the time is takes for the rock to travel to the bottom of the cliff plus the time it takes for the sound of the splash to travel back to the top. That is to say:

[itex]\displaystyle T_r + T_s = 6.9 seconds[/itex]
Tr is the travel time of the rock, and Ts is the travel time of the sound.

You should start by making an equation (you can use the equations you listed) for the travel time of the rock (Tr) as a function of distance. Then similarly, you can make an equation for the travel time of the sound(Ts).

Hope that helps. If you have more trouble, feel free to ask but it helps if you post your progress to give an idea of where you're stuck.
Thanks a whole lot. I'm not sure if this is right but this is how i solved the question :D

equation:
1) tr + ts = 6.9s
2) d = (8.3)tr + 1/2 (9.8)(tr2)
3) d = 330ts

Solution:
(8.3)tr + 1/2 (9.8)(tr2) = 330ts
(8.3)tr + 1/2 (9.8)(tr2) = 330 (6.9 - tr)
(8.3)tr + 1/2 (9.8)(tr2) = 2277 - 330tr
0 = 4.9tr2 + 338.3t - 2277
tr = (-b ± *square root*b2 - 4ac)/2a
tr = (-338.3 ± *square root*338.32 - 4(4.9)(-2277))/2(4.9)
tr = + 6.1779s OR tr = -75.21s

d = (8.3)tr + 1/2 (9.8)(tr2)
d = (8.3) (6.17) + 1/2 (9.8)(6.172)
d = 237.74861 2 SIG DIG
d = 240 m
 

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