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August 26th, 2013, 04:15 PM   #1
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Height of a ball

Height of a ball after t seconds is given by

Find the average velocity from t = 2 to t = 2.1


How do I approach this?
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August 26th, 2013, 06:11 PM   #2
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Calculate (y(2.1) - y(2))/(2.1 - 2) = -25.6.
The vertical component of the required average velocity is 25.6 height units per second, directed downwards.
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August 27th, 2013, 01:08 AM   #3
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Re: Height of a ball

Solution: Given height of the ball after t second y(t)=40t-16 t
at t=2 ,y(2)=40*2-16*(2)
y(2)=16
at t=2.1, y(2.1)=40*(2.1)-16(2.1)
y(2.1)=13.44
average velocity =[y(2.1)-y(2)]/[2.1-2]
average velocity = -25.6
therefore average velocity is 25.6,-ve sign shows direction downwards
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August 27th, 2013, 01:57 AM   #4
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Re: Height of a ball

average velocity = (final velocity + initial velocity)/2

is this valid only for constant acceleration ?
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August 27th, 2013, 01:59 AM   #5
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Re: Height of a ball

Quote:
Originally Posted by MATHEMATICIAN
average velocity = (final velocity + initial velocity)/2

is this valid only for constant acceleration ?
I think so yes. But we don't know the final nor the initial velocity.
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August 27th, 2013, 04:03 AM   #6
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Re: Height of a ball

Quote:
Originally Posted by MATHEMATICIAN
average velocity = (final velocity + initial velocity)/2

is this valid only for constant acceleration ?
Yes, only for constant acceleration. It might be more natural first to calculate the velocity by differentiating:

v(t) = 40 - 32t

v(2) = -24 and v(2.1) = -27.2
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