September 30th, 2017, 06:06 PM  #1 
Member Joined: Apr 2017 From: New York Posts: 38 Thanks: 2  velocity
If a ball is thrown vertically upward with a velocity of 144 ft/s, then its height after t seconds is s = 144t − 16t². a) What is the velocity of the ball when it is 320 ft above the ground on its way up? (Consider up to be the positive direction.) b) What is the velocity of the ball when it is 320 ft above the ground on its way down? Can you teach me how questions of this kind can be solved? Thanks. Last edited by skipjack; October 1st, 2017 at 01:43 PM. 
September 30th, 2017, 09:50 PM  #2 
Senior Member Joined: Sep 2015 From: Southern California, USA Posts: 1,488 Thanks: 749 
$s(t) = 144t 16t^2$ a) $v(t) = \dfrac{ds}{dt} = 14432t$ now we need to determine at what time(s) the ball is 320 ft above the ground. $320 = 144t  16t^2$ $16t^2 144t + 320 = 0$ $t^2  9t + 20 = 0$ $(t5)(t4) = 0$ $t = 4s,~5s$ It should be clear that $t=4s$ corresponds with it being on its way up. so $v(4) = 14432(4) = 144  128 = 16ft/s$ b) Conveniently, we've already solved for $t$ corresponding to $s=320$ and it being on its way down. We did this in (a) and saw it to be $t=5s$ $v(5) = 144  32(5) = 144160 = 16 ft/s$ Last edited by skipjack; October 1st, 2017 at 01:53 PM. 
October 1st, 2017, 07:32 AM  #3 
Member Joined: Apr 2017 From: New York Posts: 38 Thanks: 2 
Thank you very much Romsek, that helped a lot. what I was doing was solving for t find t=4, 5 but I was plugging in into original equation (144t16t^2) instead of the derivative of the function. thanks thanks thanks 
October 1st, 2017, 09:32 AM  #4 
Member Joined: Apr 2017 From: New York Posts: 38 Thanks: 2 
this is great thanks. how about finding the maximum height the particle can reach? 
October 1st, 2017, 09:44 AM  #5 
Senior Member Joined: Sep 2015 From: Southern California, USA Posts: 1,488 Thanks: 749  
October 1st, 2017, 11:08 AM  #6 
Member Joined: Apr 2017 From: New York Posts: 38 Thanks: 2 
velocity is zero at the maximum height. I equalize the f'(x) to zero and solve for the x and I see that t=4.5s which is the maximum height occurs at 4.5s so I plug it in my f(x) to see that the height is? is that so? 
October 1st, 2017, 12:05 PM  #7 
Senior Member Joined: Sep 2015 From: Southern California, USA Posts: 1,488 Thanks: 749  

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