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Mathematics 15 Online
OpenStudy (anonymous):

A projectile is fired directly upward with an initial velocity 144 ft/secs. Its height s(t) in feet above the ground after t seconds is given by s(t) = 144 - 16t^2. a) what are the velocity and acceleration after t sec? b) after 3 sec? c) What is the maximum height? d) when does it strike the ground?

OpenStudy (anonymous):

Ah, initial conditions..:-)

OpenStudy (amistre64):

this is wrong

OpenStudy (amistre64):

s(t) = 144 - 16t^2

OpenStudy (amistre64):

should be: s(t) = 144t - 16t^2

OpenStudy (anonymous):

yes, he did the v(t) and asked a question.....

OpenStudy (anonymous):

This is the whole question that 'm talking about...

OpenStudy (amistre64):

i know, but initial velocity is not the same as initial height in the equation

OpenStudy (amistre64):

s(t) = Gt^2 + Vt + H

OpenStudy (amistre64):

s(t) = -16t^2 +144t + 0

OpenStudy (anonymous):

good eye!

OpenStudy (amistre64):

bad back tho ;)

OpenStudy (anonymous):

lol!

OpenStudy (amistre64):

s(t) = -16t^2 +144t v(t) = -32t + 144 a(t) = -32

OpenStudy (amistre64):

v(3) = -32(3) + 144 = -96 + 144 a(3) = -32

OpenStudy (anonymous):

morning joe don't forget maximum height right?

OpenStudy (amistre64):

the max height is at v(t) = 0 v(t) = -32t + 144 = 0; when t = 144/32 s(9/2) = -16(9/2)^2 +144(9/2)

OpenStudy (anonymous):

SCHOOL!!! @everyone morning :)

OpenStudy (amistre64):

it strikes that ground again after another 9/2 secs so: 2(9/2) secs

OpenStudy (amistre64):

see how easy that is when you give all the info to begin with ;)

OpenStudy (amistre64):

any thoughts? or corrections?

OpenStudy (anonymous):

where is your spirit of adventure!? doing a problem with half the info needed is exciting and fun! lol

OpenStudy (amistre64):

oh it is... it is. but, there are times in the morning, you know, when the sun is barely peeking over the water crestnuts, and you gotta wonder to yourself, where did I leave me keys?

OpenStudy (anonymous):

lolol

OpenStudy (anonymous):

what is maximum height?

OpenStudy (amistre64):

s(9/2) = -16(9/2)^2 +144(9/2)

OpenStudy (amistre64):

the max height is when the thing stops moving up and starts to move down; so its velocity is at 0 v(t) = -32t + 144 = 0 when t = 144/32 = 9/2

OpenStudy (anonymous):

oh okay that's right thank you very much....

OpenStudy (amistre64):

youre are quite welcome :)

OpenStudy (amistre64):

im outta stickers but I got some of these: ♫♫♫

OpenStudy (anonymous):

cool u r truly a legend hehehe.

OpenStudy (anonymous):

can is ask another question? cause i'm going to have my recitation tommorow?

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