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Stone is released from rest at top of a vertical cliff: Height of cliff, penetration force

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Q1. A stone of mass 80 grams is released from rest at the top of a vertical cliff. After falling for 3 seconds, it reaches the ground and penetrates 9 cm into the ground.

Calculate:

1. the height of the cliff

2. the average force resisting the penetration

(assume gravitational acceleration g=10m/sec2)

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Solution Summary

A stone released from rest on top of a vertical cliff is analyzed. With good explanations and calculations, the problem is solved.

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Solution. Assume that the height of the cliff is H(m)and the average force resisting the penetration is F (N). Denote the mass of the stone by m, so m=0.08kg. ...

Solution provided by:
Changping Wang, MA
Education
  • BSc , Wuhan Univ. China
  • MA, Shandong Univ.
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  • "Your solution, looks excellent. I recognize things from previous chapters. I have seen the standard deviation formula you used to get 5.154. I do understand the Central Limit Theorem needs the sample size (n) to be greater than 30, we have 100. I do understand the sample mean(s) of the population will follow a normal distribution, and that CLT states the sample mean of population is the population (mean), we have 143.74. But when and WHY do we use the standard deviation formula where you got 5.154. WHEN & Why use standard deviation of the sample mean. I don't understand, why don't we simply use the "100" I understand that standard deviation is the square root of variance. I do understand that the variance is the square of the differences of each sample data value minus the mean. But somehow, why not use 100, why use standard deviation of sample mean? Please help explain."
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