anonymous
  • anonymous
***I WILL GIVE BEST ANSWER!***HEYY! CAN SOMEONE PLEASE HELP ME WITH THIS QUESTION!? I HAVE A TEST COMING UP AND I AM SOO CONFUZZLED! PLEASE INCLUDE THE NUMBERS AND SOLUTIONS IN ANSWER! (please dont round)
Physics
katieb
  • katieb
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anonymous
  • anonymous
anonymous
  • anonymous
The question is what is the height of where the collision takes place!
RANE
  • RANE
r u given the answer ?

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anonymous
  • anonymous
sadly, no :(
RANE
  • RANE
ahan well i dis q's like this last year , i think i remember it
anonymous
  • anonymous
really!? how do u do it!?
RANE
  • RANE
i think so, let me try first
anonymous
  • anonymous
okay! thankk youu!
anonymous
  • anonymous
i get a few tries to see if i got the question right so if you give me an answer i could put it in and see if youre right. and then you could explain it! but you cant round on anything! and this is only if youre willing to help me! sorry
RANE
  • RANE
ok i did it and i got 91.6m but im not sure if thats the answer
anonymous
  • anonymous
did you round anything?
RANE
  • RANE
nop
RANE
  • RANE
was i suppose to?
anonymous
  • anonymous
nope! okay let me check
anonymous
  • anonymous
nope its wrong :(
RANE
  • RANE
whts the answer then?
anonymous
  • anonymous
idk i can put in answers and it will tell me if its right or wrong
RANE
  • RANE
idk then i cant figure out any other way to do it
anonymous
  • anonymous
its okay thanks anyways!
RANE
  • RANE
np:) but dont u have nay clue wht could be the answer around ?
anonymous
  • anonymous
absolutely no idea lol
RANE
  • RANE
have u done q's like that before?
anonymous
  • anonymous
not as hard!
anonymous
  • anonymous
this one is really complex
RANE
  • RANE
sure it is
RANE
  • RANE
maybe he can help @UnkleRhaukus
anonymous
  • anonymous
YES PLEASE @UnkleRhaukus lol
anonymous
  • anonymous
|dw:1380448186791:dw| You know Vx and d. So you can calculate time t. Then you can find what the AP is by using \[\left| AP \right| =\frac{ g*t ^{2} }{ 2 }\] Then you can calculate BP easily. |BP|=|AB|-|AP|
anonymous
  • anonymous
For target:\[h_t=h_0-1/2g t^2\]For bullet\[h_b=v_0\sin(\theta)t-1/2g t^2\]At time t_c both bodies collide:\[h_b=h_t \rightarrow v_0\sin(\theta)t_c-1/2g t_c^2=h_0-1/2g t_c^2 \rightarrow t_c=\frac{ h_0 }{ v_0\sin(\theta) }\]and at that moment, target is at:\[h_t(t_c)=h_0-\frac{ 1 }{ 2 }g \left( \frac{ h_0 }{ v_0\sin(\theta) } \right)^2=91.7-4.9·\left( \frac{ 91.7 }{ 109·\sin(48.4)} \right)^2\]
anonymous
  • anonymous
and during time t_c, bullet has to cover distance "d". Let us see:\[x(t=t_c)=v_0\cos(\theta)t_c=v_0\cos(\theta)\frac{ h_0 }{ v_0\sin(\theta) }=h_0/\tan(\theta)=d\]

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