Question

***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)

Answer 1

The question is what is the height of where the collision takes place!

Answer 2

r u given the answer ?

Answer 3

sadly, no :(

Answer 4

ahan well i dis q's like this last year , i think i remember it

Answer 5

really!? how do u do it!?

Answer 6

i think so, let me try first

Answer 7

okay! thankk youu!

Answer 8

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

Answer 9

ok i did it and i got 91.6m but im not sure if thats the answer

Answer 10

did you round anything?

Answer 11

nop

Answer 12

was i suppose to?

Answer 13

nope! okay let me check

Answer 14

nope its wrong :(

Answer 15

whts the answer then?

Answer 16

idk i can put in answers and it will tell me if its right or wrong

Answer 17

idk then i cant figure out any other way to do it

Answer 18

its okay thanks anyways!

Answer 19

np:) but dont u have nay clue wht could be the answer around ?

Answer 20

absolutely no idea lol

Answer 21

have u done q's like that before?

Answer 22

not as hard!

Answer 23

this one is really complex

Answer 24

sure it is

Answer 25

maybe he can help @UnkleRhaukus

Answer 26

YES PLEASE @UnkleRhaukus lol

Answer 27

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|

Answer 28

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:\[h_t(t_c)=h_0-\frac{ 1 }{ 2 }g \left( \frac{ h_0 }{ v_0\sin(\theta) } \right)^2=91.7-4.9·\left( \frac{ 91.7 }{ 109·\sin(48.4)} \right)^2\]

Answer 29

and during time t_c, bullet has to cover distance "d". Let us see:\[x(t=t_c)=v_0\cos(\theta)t_c=v_0\cos(\theta)\frac{ h_0 }{ v_0\sin(\theta) }=h_0/\tan(\theta)=d\]