My Math Forum Height of a ball

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 August 26th, 2013, 04:15 PM #1 Senior Member   Joined: Apr 2013 From: Ramallah, Palestine Posts: 349 Thanks: 0 Height of a ball Height of a ball after t seconds is given by $y(t)= 40t - 16t^2$ Find the average velocity from t = 2 to t = 2.1 How do I approach this?
 August 26th, 2013, 06:11 PM #2 Global Moderator   Joined: Dec 2006 Posts: 20,629 Thanks: 2077 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.
 August 27th, 2013, 01:08 AM #3 Newbie   Joined: May 2013 Posts: 16 Thanks: 0 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
 August 27th, 2013, 01:57 AM #4 Math Team     Joined: Jul 2013 From: काठमाडौं, नेपाल Posts: 884 Thanks: 61 Math Focus: सामान्य गणित Re: Height of a ball average velocity = (final velocity + initial velocity)/2 is this valid only for constant acceleration ?
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.

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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