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September 30th, 2017, 07:06 PM   #1
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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 02:43 PM.
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September 30th, 2017, 10:50 PM   #2
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$s(t) = 144t -16t^2$

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 02:53 PM.
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October 1st, 2017, 08:32 AM   #3
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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
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October 1st, 2017, 10:32 AM   #4
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this is great thanks.

how about finding the maximum height the particle can reach?
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October 1st, 2017, 10:44 AM   #5
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Quote:
Originally Posted by Leonardox View Post
this is great thanks.

how about finding the maximum height the particle can reach?
what is the velocity at maximum height?
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October 1st, 2017, 12:08 PM   #6
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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?
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October 1st, 2017, 01:05 PM   #7
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Quote:
Originally Posted by Leonardox View Post
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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