Solve Monkey Dropping Coconut Problem on a Sunny Day

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The discussion revolves around solving a physics problem involving a monkey dropping a coconut from a tree while factoring in the time it takes for sound to travel. The key point is that the time from the coconut's release to impact is 190% longer than the time it takes for the sound of the impact to reach the observer. Participants clarify the relationships between time, height, and distance using trigonometric and kinematic equations. The height of the tree can be calculated using the formula y = 4.9t², and the horizontal distance can be derived from the speed of sound. The original poster seeks guidance on the formulas and methods to approach similar problems in the future.
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Hi,
I just joined today and would like help with a question. Here it is:

It is a sunny day and you are lying on a beach . Some distance away across level gropund, is a tall coconut palm. As you watch, a monkey in that tree drops a coconut(from rest) onto hard ground below. The time elapsed from the moment he releases the coconut until it hits the pavement is 190% longer than the time it takes for the impact to reach you. The angle (measured with respect to the ground ) at which you are viewing the monkey when he drops the nut is 5 degrees. Assumimg the speed of sound is constant 343m/s, from what height did the monkey drop the nut?
Thanks
 
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time for sound to reach you = (1.90t + t) =2.90t
time for coconut to fall = t
height of tree = y
distance from tree = x = y/(tan 5)
 
bukky, if this is homework or course work you should show us what you have tried on this. (If it isn't, this is the wrong forum.)

Chi Meson, did you misread the question? Bukky said ". The time elapsed from the moment he releases the coconut until it hits the pavement is 190% longer than the time it takes for the impact to reach you." (I presume the "impact" is the sound of the coconut hitting the pavement.) You say "time for sound to reach you = (1.90t + t) =2.90t". If t is the time it takes for the coconut to hit the ground, then the time for the sound to reach you is t/1.90.
Letting x be the horizontal distance to the bottom of the tree (where the coconut hits) then x= 343t/1.9= 180.5t. Letting y be the height of the tree, then y= 4.9t2. You can solve
tan(5)= y/x= (4.9/180.5)t= 0.027t for t and then find y.
 
HallsofIvy said:
bukky, if this is homework or course work you should show us what you have tried on this. (If it isn't, this is the wrong forum.)

Chi Meson, did you misread the question? Bukky said ". The time elapsed from the moment he releases the coconut until it hits the pavement is 190% longer than the time it takes for the impact to reach you." (I presume the "impact" is the sound of the coconut hitting the pavement.) You say "time for sound to reach you = (1.90t + t) =2.90t". If t is the time it takes for the coconut to hit the ground, then the time for the sound to reach you is t/1.90.
Letting x be the horizontal distance to the bottom of the tree (where the coconut hits) then x= 343t/1.9= 180.5t. Letting y be the height of the tree, then y= 4.9t2. You can solve
tan(5)= y/x= (4.9/180.5)t= 0.027t for t and then find y.

It is an assignment and we were( class) given any hunts on how to solve the problem.
I have attempted the problem numerous times. I was using the formula
v1-vo=2ax. I do have a question? how did you obtain 4.9t? is there a formula I could use to approach such a question in the future? I want to make sure I understand it. Rather than just obtaining the answer.
Thanks for your help
 
Chi Meson said:
time for sound to reach you = (1.90t + t) =2.90t
time for coconut to fall = t
height of tree = y
distance from tree = x = y/(tan 5)
Thanks for your help!
 
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