PHY 231
Your 'cq_1_08.2' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
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You should enter your answers using the text editor or word processor. You will then copy-and-paste it into the box below, and submit.
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A ball is tossed upward at 15 meters / second from a height of 12 meters above the ground. Assume a uniform downward acceleration of 10 m/s^2 (an approximation within 2% of the 9.8 m/s^2 acceleration of gravity). How high does it rise and how long does it take to get to its highest point? answer/question/discussion:
It takes a second and a half for the ball to reach its highest point at 3.75 m above the initial height, or 15.75 m.
Calculations:
Velocity = 15 - 10t
0 m ( highest point ) = 15 - 10t
t ( time in seconds ) = 1.5
y ( position ) = 10t - 5t^2
y ( position ) = 3.75 m
3.75 m + 12.00 m = 15.75 m
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• How fast is it then going when it hits the ground, and how long after the initial toss does it first strike the ground?
answer/question/discussion: ->->->->->->->->->->->-> :
It takes about 2.84 seconds to hit the ground at -13.43908891 m/s.
Calculations:
y ( position ) = 10t - 5t^2 + 12
0 = -5t^2 + 10t + 12
Quadratic Equation ( t = -b+/- ( sqrt.( b^2 -4ac ))/ ( 2a ) )
t = 2.843908891 seconds ≈ 2.84 seconds
Velocity = 15 - 10t
= 15 - 10( 2.843908891 s )
= -13.43908891 m/s
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• At what clock time(s) will the speed of the ball be 5 meters / second?
answer/question/discussion: ->->->->->->->->->->->-> :
One second after the toss, the speed of the ball will be 5 m/s.
Calculations:
Velocity = 15 - 10t
5 m/s = 15 - 10t
t = 1
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• At what clock time(s) will the ball be 20 meters above the ground?
• How high will it be at the end of the sixth second?
answer/question/discussion: ->->->->->->->->->->->-> :
The ball will never be 20 meters above ground; its highest point is 15.75 meters as stated above. At the end of the sixth second, it will be on the ground, since the result is below 0. It might be bouncing/rolling, making contact with the ground.
Calculations:
y ( position ) = 10t - 5t^2 + 12
20 = 10t -5t^2 + 12
0 = -5t^2 + 10t - 8
Quadratic Equation ( t = -b+/- ( sqrt.( b^2 -4ac ))/ ( 2a ) )
t = Imaginary Number ( Does not exist...It will never get 20 m above ground)
y ( position ) = 10t - 5t^2 + 12
y = 10(6) - 5(6)^2 + 12
y = 60 - 5 * 36 + 12
y = 60 -180 + 12
y = -108 ( Below Ground Level Theoretically )
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I think I solved these problems better than my last set of seed questions from assignment 8, using functions versus drawings and intervals.
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Copy and paste your work into the box below and submit as indicated:
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35 Minutes
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