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Water mechanics homework

by Badrakhandama
Tags: angle, mechanics, speed, velocity, voltage
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Badrakhandama
#1
Jan30-10, 04:29 AM
P: 25
A cliff diver takes off horizontally from a cliff face 26.5 metres above the surface of the water. In the course of his flight, he travels 8 metres forwards, of which all but the last 1.5 metres of horizontal motion is above rock. In other words, he travels 6.5 metres in the air horizontally, before he falls and travels the remaining 1.5 metres.

Questions:
1) How long is he in the air?
s = 1/2 (g) (t^2), so time =2.3 seconds

2) How long is he OVER the water during the flight?

I assume this is asking how long he travels at the 6.5 metres?

I have no idea what equation to use or where to start - all i know is that the answer is 0.44 seconds...HELP!
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torquil
#2
Jan30-10, 07:24 AM
P: 641
Quote Quote by Badrakhandama View Post
A cliff diver takes off horizontally from a cliff face 26.5 metres above the surface of the water. In the course of his flight, he travels 8 metres forwards, of which all but the last 1.5 metres of horizontal motion if above rock. In other words, he travels 6.5 metres in the air, horizontally
I don't understand this exercise. Isn't "he travels 8 metres forwards" and "he travels 6.5 metres in the air, horizontally" a contradiction?

I interpreted it as 6.5m of the horizontal motion was above rock since the rock face is not vertical, and the last 1.5m is above water. So "he travels 6.5 metres in the air, horizontally" would be incorrect, no?

Torquil
Badrakhandama
#3
Jan30-10, 07:55 AM
P: 25
Hi - It is my fault - I have phrased it poorly.

The Diver takes off and travels forward 8 metres, by the time he has hit the water - however, 6.5 metres of the 8 is covered above the rock. The other 1.5 metres is covered as he is now falling down past the rock.

torquil
#4
Jan30-10, 08:06 AM
P: 641
Water mechanics homework

Ok, then I understand more. You have already found the amount of time. You can assume that the horizontal component of the velocity is constant since there are no horizontal forces acting (neglecting wind resistance), so you can figure out the horizontal speed.

From that you should be able to find the answer to the second question.

Torquil
Badrakhandama
#5
Jan30-10, 08:24 AM
P: 25
I agree - but I do not appear to get the answer of 0.44 seconds.

This is how I did it -

If t = 2.3 seconds

Speed = d/t =8/2.3 = 3.5 metres per second

Assuming velocity remains the same (which it does), then to travel 6.5 metres will take :
T = d/s = 6.5/3.5 = 1.87 seconds.

What am i doing wrong?
torquil
#6
Jan30-10, 08:32 AM
P: 641
I don't think you did anything wrong apart from misinterpret your result. Your time is the time above rock, so you want the difference between those time values.

Torquil
Badrakhandama
#7
Jan30-10, 08:37 AM
P: 25
Thank you! That gives the correct answer.

But I have one question - If it takes 2.3 seconds to travel 8 metres, and takes 1.87 seconds to travel 6.5 metres - then why is the answer not 1.87? Since the time above the rock is the time taken to travel 6.5 metres, right?
Badrakhandama
#8
Jan30-10, 08:40 AM
P: 25
Wait - maybe I am misinterpreting the whole question - when is says ' how long over the water,' what does that mean?


Thank You
torquil
#9
Jan30-10, 08:45 AM
P: 641
It means that they want the amount of time during which he has water directly beneath him. So that would be the time it takes to move the last 1.5m.

Torquil
Badrakhandama
#10
Jan30-10, 08:52 AM
P: 25
Ok, but he has water beneath him for all 8 metres.

The question does not say that he runs 6.5 metres and then falls, it says he travels 8 metres in the air...

So i am not sure
torquil
#11
Jan30-10, 09:00 AM
P: 641
I think it is like this: He jumps out from the rock face. the rock face is not vertical, so he has to jump at least 6.5m out to awaid being smashed on the rocks. He jumps out 8m to be sure to not hit the rocks.

So in total: 8m horizontally. 6.5 of them while with rocks underneat, and the other 1.5 with water underneath.

Torquil
Badrakhandama
#12
Jan30-10, 09:10 AM
P: 25
Ah! That makes much more sense! Thank You

Finally, make I ask you - how could you calculate 'the angle which the diver makesm to the vertical, upon entry to the water?'
Badrakhandama
#13
Jan30-10, 09:11 AM
P: 25
Oh! That seems right - thank you very much!

One last question - how do you calculate 'the angle which the diver makes, to the vertical, upon entry to the water' ???

Thank You, again
torquil
#14
Jan30-10, 09:17 AM
P: 641
No problem, you already know the horizontal speed upon impact. Since you know the total time you can also calculate vertical speed. After that you should be ble to find the angle using a trigonometric function.

Torquil
Badrakhandama
#15
Jan30-10, 09:24 AM
P: 25
Yes, I have got the horizontal speed to be 3.5 metres per second
The vertical speed is 11.5 metres per second

But what trignometric function? I do not know of any which can predict the vertical entry angle of a perabola
Badrakhandama
#16
Jan30-10, 09:39 AM
P: 25
?????
torquil
#17
Jan30-10, 09:39 AM
P: 641
I'll only tell you that it is sin, cos, or tan. You can check their definitions here:

http://en.wikipedia.org/wiki/Trigonometric_function

Good luck!

Torquil


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