## natasha.aries 2 years ago ***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)

1. natasha.aries

2. natasha.aries

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

3. RANE

r u given the answer ?

4. natasha.aries

5. RANE

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

6. natasha.aries

really!? how do u do it!?

7. RANE

i think so, let me try first

8. natasha.aries

okay! thankk youu!

9. natasha.aries

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

10. RANE

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

11. natasha.aries

did you round anything?

12. RANE

nop

13. RANE

was i suppose to?

14. natasha.aries

nope! okay let me check

15. natasha.aries

nope its wrong :(

16. RANE

17. natasha.aries

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

18. RANE

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

19. natasha.aries

its okay thanks anyways!

20. RANE

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

21. natasha.aries

absolutely no idea lol

22. RANE

have u done q's like that before?

23. natasha.aries

not as hard!

24. natasha.aries

this one is really complex

25. RANE

sure it is

26. RANE

maybe he can help @UnkleRhaukus

27. natasha.aries

28. oksuz_

|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|

29. CarlosGP

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

30. CarlosGP

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