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 September 30th, 2017, 06:06 PM #1 Member   Joined: Apr 2017 From: New York Posts: 65 Thanks: 6 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: USA Posts: 1,980 Thanks: 1027 $s(t) = 144t -16t^2$ a) $v(t) = \dfrac{ds}{dt} = 144-32t$ now we need to determine at what time(s) the ball is 320 ft above the ground. $320 = 144t - 16t^2$ $16t^2 -144t + 320 = 0$ $t^2 - 9t + 20 = 0$ $(t-5)(t-4) = 0$ $t = 4s,~5s$ It should be clear that $t=4s$ corresponds with it being on its way up. so $v(4) = 144-32(4) = 144 - 128 = 16ft/s$ 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 $t=5s$ $v(5) = 144 - 32(5) = 144-160 = -16 ft/s$ 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: 65 Thanks: 6 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 (144t-16t^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: 65 Thanks: 6 this is great thanks. how about finding the maximum height the particle can reach?
October 1st, 2017, 09:44 AM   #5
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 Originally Posted by Leonardox this is great thanks. how about finding the maximum height the particle can reach?
what is the velocity at maximum height?

 October 1st, 2017, 11:08 AM #6 Member   Joined: Apr 2017 From: New York Posts: 65 Thanks: 6 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
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 Originally Posted by Leonardox 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?
yessir.

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