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steposvetlana [31]
3 years ago
10

A diver running 1.8 m/s dives out horizontally from the edge of a vertical cliff and reaches the water below 2 s later. How high

was the cliff and how far from its base did the diver hit the water?
work needed

Mathematics
1 answer:
DIA [1.3K]3 years ago
3 0
Refer to the diagram shown below.

h =  the height of the cliff.
d = the distance of the landing spot from the base of the cliff.
Assume that air resistance may be neglected, and g = 9.8 m/s².

The horizontal distance traveled in 2 seconds is
d = (1.8 m/s)*(2 s) = 3.6 m

The initial vertical velocity component is zero.
Therefore the vertical height traveled before landing on the water obeys the equation
h = (1/2)*(9.8 m/s²)*(2 s)² = 19.6 m

Answers:
The height of the cliff is 19.6 m
The distance of the landing spot from the base of the cliff is 3.6 m

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p_v =2*P(t_{(1699)}>1.971)=0.049  

If we compare the p value and the significance level given \alpha=0.1 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is different from 634 at 10% of signficance.

Step-by-step explanation:

Data given and notation  

\bar X=669 represent the sample mean

s=732 represent the sample standard deviation

n=1700 sample size  

\mu_o =68 represent the value that we want to test

\alpha=0. represent the significance level for the hypothesis test.  

t would represent the statistic (variable of interest)  

p_v represent the p value for the test (variable of interest)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the mean is different from 634, the system of hypothesis would be:  

Null hypothesis:\mu = 634  

Alternative hypothesis:\mu \neq 634  

If we analyze the size for the sample is > 30 but we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}  (1)  

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".  

Calculate the statistic

We can replace in formula (1) the info given like this:  

t=\frac{669-634}{\frac{732}{\sqrt{1700}}}=1.971    

P-value

The first step is calculate the degrees of freedom, on this case:  

df=n-1=1700-1=1699  

Since is a two sided test the p value would be:  

p_v =2*P(t_{(1699)}>1.971)=0.049  

Conclusion  

If we compare the p value and the significance level given \alpha=0.1 we see that p_v so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is different from 634 at 10% of signficance.

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