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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 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)=40t16 t▓ at t=2 ,y(2)=40*216*(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)]/[2.12] 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: рдХрд╛рдардорд╛рдбреМрдВ, рдиреЗрдкрд╛рд▓ Posts: 884 Thanks: 61 Math Focus: рд╕рд╛рдорд╛рдиреНрдп рдЧрдгрд┐рдд  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  
Senior Member Joined: Mar 2012 From: Belgium Posts: 654 Thanks: 11  Re: Height of a ball Quote:
 
August 27th, 2013, 04:03 AM  #6  
Senior Member Joined: Jun 2013 From: London, England Posts: 1,316 Thanks: 116  Re: Height of a ball Quote:
v(t) = 40  32t v(2) = 24 and v(2.1) = 27.2  

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