IsTim
  • IsTim
A medieval prince trapped in a castle wraps a message around a rock and throws it from the top of the castle with an initial velocity of 12m/s[42 degrees of above the horizontal]. The rock lands just on the far side of the castle's moat, at a level 9.5m below the initial level. Determine the rock's time of flight.
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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King
  • King
hmmm...do u hav the answer..?
IsTim
  • IsTim
I don't know how to get it, but the answer is 2.4s.
King
  • King
ok...wait.....

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King
  • King
is a diagram given?
IsTim
  • IsTim
Yes!
King
  • King
can u post it......it will be very helpful.....
IsTim
  • IsTim
|dw:1327933563123:dw| IT's a lil rough
King
  • King
i am gettin 1.6
IsTim
  • IsTim
?
King
  • King
i got 2.4!!!!!!
King
  • King
istim u there?
IsTim
  • IsTim
How?
King
  • King
see for horizontal the time of flight is 2usintheta/g and fr vertical it is usintheta/g......so we need total T.O.F. so adding we get 3usintheta/g.....now substitute values..... 3*12*sin42/9.8 36*0.6691/9.8 24.0876/9.8 2.4s!
IsTim
  • IsTim
fr? TOF?
King
  • King
time of flight[T.O.F]
IsTim
  • IsTim
and fr?
King
  • King
for horizontal motion and vertical motion[fr a projectile there always 2 types of motion horizontal and vertical]
King
  • King
ok...got it?
IsTim
  • IsTim
LEt me review your statement. It's a bit hard to understand....
IsTim
  • IsTim
wats u in 2usintheta/g
IsTim
  • IsTim
And how did you get those formulas>?
King
  • King
u is initial velocity
King
  • King
and those formulas can be derived
King
  • King
but fr now just remember them
IsTim
  • IsTim
Sorry. What do I derive those formulas from?
King
  • King
that is super long....just remember them or google it
IsTim
  • IsTim
What should i be googling?
King
  • King
the derivation fr the formulas....range,max.height,t.o.f,etc/
IsTim
  • IsTim
\[n _{1}\sin \theta_{1}=n _{2}\sin \theta_{2}\]
IsTim
  • IsTim
is it that one?
IsTim
  • IsTim
hello?
King
  • King
http://en.wikipedia.org/wiki/Projectile_Motion
King
  • King
ok?
IsTim
  • IsTim
I never learnt that formula in class. I'm doing an exam, so what is my best alternative?
King
  • King
idk....
IsTim
  • IsTim
Any opinions mashy?
anonymous
  • anonymous
wait lemme read it again.. i was just reading the comments :D
IsTim
  • IsTim
k
anonymous
  • anonymous
nah.. i can't give a better solution sorry.. you need to know the dynamics of a parabolic tragectory!
IsTim
  • IsTim
Bugger. Ok. I'll look thru my textbook more thoroughly then.
IsTim
  • IsTim
There isn't an example with that formula in it. Argh
IsTim
  • IsTim
The initial Velocity is the component of the original velocity right?
IsTim
  • IsTim
wait, why is it multiplied by 3?
King
  • King
READ MY COMMENT!
IsTim
  • IsTim
eh. which one?
anonymous
  • anonymous
|dw:1327927068621:dw| looking in the vertical direction,we have usin42 and acceleration g downwards at point A,final velocity for path till A becomes o so apply equations for uniform acceleration v-u/t=a the SAME TIME is taken for stone to come to the same height so total time for 1st part is 2*usin42/g THIA GIVES US T1 the rock doesnt stop once it reaches the height from which the prince threw it does it? so we have a 2nd part in this problem |dw:1327927440627:dw| when it reaches point b(from where prince originally threw) at the same level, THE VELOCITY OF ROCK BECOMES SAME AS THE VELOCITY WITH WHICH PRINCE THREW that is usin42 now again apply equations for uniform acceleration |dw:1327927625023:dw| T1+T2 is the total time taken by rock! u have any doubt u may ask

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