Kinematics: Two-Dimensional Motion

self check Projectile Motion Self-Check

Now, you try. Complete the self-check activity by working out the problems, and then clicking on the question to review the explanation.

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An athlete executing a long jump leaves the ground at a 28° angle and travels 7.80 m. (a) What was the takeoff speed? [HINT: Before attempting this problem, look at example 3- in your textbook about finding "Level Horizontal Range."
Given:
angle equals twenty eight degrees; Delta x equals seven point eight zero meters
Find:
Vector v		 Equation:
Trig functions

Y equals y not plus v not y times t plus one half times g times t squared

X equals x not plus v not x times t
Solution:
Using the level horizontal range equation from the textbook example (you can always derive this, but that will take up too much space here) solve for Vector v.

R equals the quantity of v not squared times sine of 2 theta end quantity over g; R times g divided by sine of two theta equals the quantity of v not squared times sine of 2 theta end quantity over sine of two theta; V not equals square root of the quantity R times g over sine of two theta end quantity equals square root of the quantity of seven point eight zero times nine point eight over sine of the quantity 2 times twenty eight degrees end quantity end quantity. Equals nine point six zero meters per second


A projectile is fired with an initial speed of 65.2 m/s at an angle of 34.5° above the horizontal on a long flat firing range. Determine:

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(a) the maximum height reached by the projectile,
Given:
v equals sixty five point two meters per second, thirty four point five degrees; V not x equals sixty five point two times cosine of thirty four point five degrees equals fifty three point seven three meters per second; V not y equals sixty five point two times sine of thirty four point five degrees equals thirty six point nine three meters per second; Y not equals zero meters
Find:
height, y		 Equation:
V sub y squared equals v not y squared plus 2 times g times delta y
Solution:
V sub y squared equals v not y squared plus 2 times g times delta y; Zero squared equals the square of thirty six point nine three meters per second plus two times negative nine point eight times the quantity y minus zero; Zero equals one thousand three hundred sixty three point eight two five minus nineteen point six times y; Nineteen point six times y equals one thousand three hundred sixty three point eight two five; Y equals sixty nine point six meters

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(b) the total time in the air,
Given:
V equals sixty five point two meters per second, thirty four point five degrees; V not x equals sixty five point two times cosine of thirty four point five degrees equals fifty three point seven three meters per second; V not y equals sixty five point two times sine of thirty four point five degrees equals thirty six point nine three meters per second; Y equals y not equals zero meters
Find:
time, t (hang time)		 Equation:
y equals y not plus v not y times t plus one half times g times t squared
Solution:
y equals y not plus v not y times t plus one half times g times t squared; zero equals zero plus thirty six point nine three times t minus four point nine times t squared; four point nine times t squared equals thirty six point nine three times t; t equals seven point five four seconds

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(c) the total horizontal distance covered (range),
Given:
V equals sixty five point two meters per second, thirty four point five degrees; V not x equals sixty five point two times cosine of thirty four point five degrees equals fifty three point seven three meters per second; V not y equals sixty five point two times sine of thirty four point five degrees equals thirty six point nine three meters per second; t equals seven point five four seconds
Find:
Delta x		 Equation:
X equals x not plus v not x times t
Solution:
X equals x not plus v not x times t; X minus x not equals v not x times t; Delta x equals fifty three point seven three meters per second times seven point five four seconds; Delta x equals four hundred five meters

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(d) and the velocity of the projectile 1.50 s after firing.
Given:
V equals sixty five point two meters per second, thirty four point five degrees; V not x equals sixty five point two times cosine of thirty four point five degrees equals fifty three point seven three meters per second; V not y equals sixty five point two times sine of thirty four point five degrees equals thirty six point nine three meters per second; Time equals one point five zero seconds
Find:
Vy after 1.50 seconds, then resultant velocity		 Equation:
V sub y equals v not y plus g times t
Trig functions, Pythagorean Theorem
Solution:
V sub y equals v not y plus g times t; V sub y equals thirty six point nine three meters per second plus negative nine point eight meters per second squared times one point five zero seconds; V sub y equals twenty two point two three meters per second

Then, find the magnitude of the resultant:

V equals square root of the quantity of twenty two point two three squared plus fifty three point seven three squared end equals fifty eight point one meters per second

And the direction of the resultant:

Tangent theta equals twenty two point two three divided by fifty three point seven three; Theta equals inverse tangent of the fraction twenty two point two three over fifty three point seven three; Theta equals twenty two point five degrees

 

self check More Practice Problems

Problem 1

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Problem 2

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Problem 3

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Problem 4

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Text book coverFor more practice, solve the following problems from "Chapter 3, Problems" at the end of the chapter. The answers are in the back of your book.

Problems # 19, 23, 27, 31, 35