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Anna71 [15]
1 year ago
14

You shoot an arrow into the air. Two seconds later (2.00 s) the arrow has gone straight upward to a height of 35.0 m above its l

aunch point. What was the arrow's initial speed? How long did it take for the arrow to first reach a height of 17.5 m above its launch point?
Physics
1 answer:
sdas [7]1 year ago
4 0

This question can be solved by using the equations of motion.

a) The initial speed of the arrow is was "9.81 m/s".

b) It took the arrow "1.13 s" to reach a height of 17.5 m.

a)

We will use the second equation of motion to find out the initial speed of the arrow.

h= v_it + \frac{1}{2}gt^2\\

where,

vi = initial speed = ?

h = height = 35 m

t = time interval = 2 s

g = acceleration due to gravity = 9.81 m/s²

Therefore,

35\ m = (v_i)(2\ s)+\frac{1}{2}(9.81\ m/s^2)(2\ s)^2\\\\v_i(2\ s)=19.62\ m\\\\v_i = \frac{19.62\ m}{2\ s}

<u>vi =  9.81 m/s</u>

b)

To find the time taken by the arrow to reach 17.5 m, we will use the second equation of motion again.

h= v_it + \frac{1}{2}gt^2\\

where,

g = acceleration due to gravity = 9.81 m/s²

h = height = 17.5 m

vi = initial speed = 9.81 m/s

t = time = ?

Therefore,

17.5 = (9.81)t+\frac{1}{2}(9.81)t^2\\4.905t^2+9.81t-17.5=0

solving this quadratic equation using the quadratic formula, we get:

t = -3.13 s (OR) t = 1.13 s

Since time can not have a negative value.

Therefore,

<u>t = 1.13 s</u>

Learn more about equations of motion here:

brainly.com/question/20594939?referrer=searchResults

The attached picture shows the equations of motion in the horizontal and vertical directions.

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Describe in a couple sentences how the index of refraction is related to the speed of light?
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7. These temperatures were recorded in Pasadena for a week in April. 87 85 80 78 83 86 90 Find each of these. (a) Mean (e) Range
slega [8]

Answer:

a) Mean = 84.14

b) Median = 85

c) Mode = no mode (since there is no variable that appears more than once in this dataset)

d) Midrange = 84

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g) Standard deviation = 3.83

Explanation:

The raw data to be processed is

87 85 80 78 83 86 90

a) Mean = (Σx)/N

The mean is the sum of variables divided by the number of variables

x = each variable

N = number of variables = 7

Mean = (87+85+80+78+83+86+90)/7

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b) Median is the number in the middle of the dataset when the variables are arranged in ascending or descending order.

Arranging the data in ascending order

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The number in the middle is the 4th number = 85

Median = 85

c) Mode is the variable that occurs the most in a distribution.

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d) Midrange is the arithmetic mean of the highest and lowest number in the dataset.

Mathematically,

Midrange = (Highest + Lowest)/2

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e) Range is the difference the highest and the lowest numbers in a dataset.

Range = 90 - 78 = 12

f) Variance is an average of the squared deviations from the mean.

Mathematically,

Variance = [Σ(x - xbar)²/N]

xbar = mean

Σ(x - xbar)² = (78 - 84.14)² + (80 - 84.14)² + (83 - 84.14)² + (85 - 84.14)² + (86 - 84.14)² + (87 - 84.14)² + (90 - 84.14)² = 102.8572

Variance = (102.8572)/7

Variance = 14.69

g) Standard deviation = √(variance)

Standard deviation = √(14.69)

Standard deviation = 3.83

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