# Shooting a sniper rifle

1. Feb 26, 2006

### taffman123

hello - i am a first year university student and recieved an assignment last week - one question i am having very difficulty with is a question about shooting a rifle gun. the question is

If you wanted to shoot an objec on a building that is 40 m high and and 1500 m far with a velocity of V what would be the angle? is there anyone who can get me on teh rigth track?

2. Feb 26, 2006

### Hootenanny

Staff Emeritus
Hi, sorry posted in the other thread. How do you think you should start?

3. Feb 26, 2006

### Hootenanny

Staff Emeritus
What kinematic equations do you know? Think about the horizontal component because now we have 1500m this is the easiest componet to calculate.

4. Feb 26, 2006

### taffman123

what do u mean?

5. Feb 26, 2006

### Hootenanny

Staff Emeritus
You must know some kinematic equations, $v =u +at$, for example.

6. Feb 26, 2006

### taffman123

i know 5 kinematic equations

but only 4 i can use because i do not know what v2 is

7. Feb 26, 2006

### taffman123

d = v1t + 1/2at^2 = that is the one i am primarily trying to use
and also i am thinking of using d=vt

8. Feb 26, 2006

### Hootenanny

Staff Emeritus
So, what do you know and what do you need to know? Which equation(s) statisfy the criteria?

9. Feb 26, 2006

### taffman123

with having a d = 1500m
v = 854cosx
and an unknown time it seems easy - but for me i got confused with the cosx

would my time = 1.756cosx? or would it = 1.756/cosx?

10. Feb 26, 2006

### Hootenanny

Staff Emeritus
$d = vt$ is the same as $d = ut + \frac{1}{2} at^2$ without any acceleration. Which is the case horizontally, therefore you can calculate the horizontal flight time as a function of $\theta$

11. Feb 26, 2006

### taffman123

(y axis)

d = 40m
t=?
a=-9.8m/s
v1= 854sinx

makin my equation

40=854sinxt + -4.9t^2

12. Feb 26, 2006

### Hootenanny

Staff Emeritus
Your time would be $t = \frac{1500}{\cos x}$

13. Feb 26, 2006

### taffman123

what happened to the v of 854?

14. Feb 26, 2006

### Hootenanny

Staff Emeritus
Sorry I missed the co-efficent off. My mistake

15. Feb 26, 2006

### taffman123

so when i isolate for t (vertical) would it look like

2t =Square root of (40/854sinx) X 1/-4.9 ? or did i make an isolation mistake

16. Feb 26, 2006

### Hootenanny

Staff Emeritus
I'm afraid you cannot isolate $t$, the equation is a quadratic.

17. Feb 26, 2006

### taffman123

thats what i tohught because i had a t + t^2 but i just thought if i brought the ^2 over it would be ok

18. Feb 26, 2006

### Hootenanny

Staff Emeritus
No because only one term is squared, when you root both sides you will get $t + \sqrt{t} = ...$.

19. Feb 26, 2006

### taffman123

ok - so now my equation would look like

40/854sinx * 1/-4.9 = t + t^2 and now im confused

the only thing i could do is substitute t = 1500/854cosx?

20. Feb 26, 2006

### taffman123

but having all these cosx and sinx in the equation is going to fustrate me so much