Time to Get Away from Avalanache

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SUMMARY

The discussion focuses on solving physics problems related to motion under gravity and the effects of forces on falling objects. The first problem involves calculating the time to escape an avalanche on a 30-degree slope, where the correct formula incorporates the sine of the angle to determine the effective gravitational acceleration. The second problem addresses a ball dropped from a height of 176.4 meters, factoring in a horizontal wind force of 12.0 N. The solutions provided clarify the importance of using the correct components of gravitational force when analyzing motion on an incline.

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atse1900
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Hi. Can someone help me on these problems? I don't understand how to do it.

1) In an avalanche, a mass of snow and ice high on a mountain breaks llose and starts an esseentially frictionless "ride" down the mountain on a cushion of compressed air. If you were on a 30 degrees slope and an avalanche starrted 400 m up the slope, how much time would you have to get out of the way?


sin 30 * 400 =1/2gt^2
200=1/2gt^2
400=9.81t^2
t^2=40.77
t=6.32855 : (

All I got... but it's not the same answer as in the back of the book. lol k. : ].
2) A 3.00 kg ball is dropped from the roof of a building 176.4 m high. While the ball is falling to oEart, a horizontal wind exerts a constant force of 12.0 N on it. a)How far from the building the building does the ball hit the ground? b)How long does it take to hit the ground? c) What is its speed when it hits the ground?

If someone can help me, I apperciate it. Thanks.
 
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d = 1/2 gt^2

Your d is still 400, not 400sin30. It says 400m "up the slope", so that is your d. The 30 degrees comes in when you are dealing with the g value. If you drop a ball and it falls straight down it accelerates at g. If you drop a ball down a slope of 30 degrees, it accelerates at gsin30, because it is only the vertical component that matters with gravity.

So in your question it should be 400 = .5(9.8)(sin30)t^2

t = 12.8s
 
Thank you.
 

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