Hey guys I know posting a thread asking for help on a problem isn't ideal but I'm stuck here and have absolutely nowhere to turn to. I tried yahoo answers but nobody answered there and its been around 3 hours since I posted the question. Also this seemed like the forum most suited to my dilemma.
So here's the question:
A rifle has a maximum range of 500m. (a) For what angles of elevation would the range be 350m? What is the range when the bullet leaves the rifle (b) at 14 degrees? (c) at 76 degrees?
Answer is: Doesn't matter, keep shooting all the kids in the school.
[QUOTE=united forever;25670943]Hey guys I know posting a thread asking for help on a problem isn't ideal but I'm stuck here and have absolutely nowhere to turn to. I tried yahoo answers but nobody answered there and its been around 3 hours since I posted the question. Also this seemed like the forum most suited to my dilemma.
So here's the question:
A rifle has a maximum range of 500m. (a) For what angles of elevation would the range be 350m? What is the range when the bullet leaves the rifle (b) at 14 degrees? (c) at 76 degrees?[/QUOTE]
500m(9.81m^2)x0.526(Kevlar`s constant)= 3.141592654
[editline]27th October 2010[/editline]
Make sure you use Newton's fifth law of thermodynamics for figuring out the square root
[QUOTE=VQ35HR;25670993]500m(9.81m^2)x0.526(Kevlar`s constant)= 3.141592654[/QUOTE]
HURR PI.
I actually never learned this in physics in school. Why don't you try to find a physic's formula, or study your science book. Most books will have all and more formulas then you'd ever want.
I feel this is relevant but I'm not sure. [url]http://en.wikipedia.org/wiki/Trajectory_of_a_projectile#Angle_of_reach[/url]
do your own homework op.
I might've done shit at physics at school (because I barely studied), but it's really not that hard. You just need to figure out which of the many equations you have learnt applies here and then fill in the blanks with data from the question.
As people have mentioned above, look in your textbook.
I could do it if I really felt motivated. Read some of the things on here [url]http://www.physicsclassroom.com/class/vectors/u3l2e.cfm[/url]
I've looked in the textbook I know all the motion equations and when they apply. This problem is a lot more complex than that (at least it looks like it to me). I've been wrecking physics this quarter and out of nowhere this comes along. Does anyone know what to do?
What position is the gun? Like 5m above ground or what?
[QUOTE=VQ35HR;25670993]500m(9.81m^2)x0.526(Kevlar`s constant)= 3.141592654
[editline]27th October 2010[/editline]
Make sure you use Newton's fifth law of thermodynamics for figuring out the square root[/QUOTE]
remember: BTU/hr * ft(squared) * Fahrenheit for the value
[QUOTE=united forever;25670943]Hey guys I know posting a thread asking for help on a problem isn't ideal but I'm stuck here and have absolutely nowhere to turn to. I tried yahoo answers but nobody answered there and its been around 3 hours since I posted the question. Also this seemed like the forum most suited to my dilemma.
So here's the question:
A rifle has a maximum range of 500m. (a) For what angles of elevation would the range be 350m? What is the range when the bullet leaves the rifle (b) at 14 degrees? (c) at 76 degrees?[/QUOTE]
You write the range of the rifle as a function of the firing angle. (a) involves solving the angle when f(angle) = 350m and for (b) and (c) you just input the angle into the function and get the range.
[editline]27th October 2010[/editline]
The first sentence is required in writing down the function. You probably should prove that for any muzzle velocity v you can get the maximum range by firing it at an angle of 45 degrees, though that might not be required. But you'll know then that f(45) = 500m
[editline]27th October 2010[/editline]
Well, assuming you don't need to account for air resistance
it'd be awesome if you'd tell us the height above the ground the rifle shoots at or the angle of the gun when you get 500 m.
[QUOTE=ThePuska;25671473]You write the range of the rifle as a function of the firing angle. (a) involves solving the angle when f(angle) = 350m and for (b) and (c) you just input the angle into the function and get the range.
[editline]27th October 2010[/editline]
Yea I know that 45 degrees gives you the furthest distance but I really don't know how to solve for the values. I've tried triangles/solving for velocities anything I could think of but nothing is coming together on this problem for me. And air resistance is neglected.
The first sentence is required in writing down the function. You probably should prove that for any muzzle velocity v you can get the maximum range by firing it at an angle of 45 degrees, though that might not be required. But you'll know then that f(45) = 500m
[editline]27th October 2010[/editline]
Well, assuming you don't need to account for air resistance[/QUOTE]
Yea I know that 45 degrees gives you the furthest distance but I really don't know how to solve for the values. I've tried triangles/solving for velocities anything I could think of but nothing is coming together on this problem for me. And air resistance is neglected.
also rifle is fired from ground level
If the angle for range is 45deg then 45/500*350=31.5 ,500/45*14=155.55 & 500/45*76=844.44
[editline]27th October 2010[/editline]
So do you have the velocity or any other information on it?
Because then you can use: [img]http://upload.wikimedia.org/math/7/8/a/78a86dc847fd52baa1eaeb66b697c39e.png[/img]
If there's one thing I learned from physics is to take things step by step. Start small, draw out a diagram, try eliminating some variables and do things one by one. Think through it coherently and it'll be a walk in the park.
[QUOTE=RELAXiN;25671703]If the angle for range is 45deg then 45/500*350=31.5 ,500/45*14=155.55 & 500/45*76=844.44
[editline]27th October 2010[/editline]
So do you have the velocity or any other information on it?
Because then you can use: [img_thumb]http://upload.wikimedia.org/math/7/8/a/78a86dc847fd52baa1eaeb66b697c39e.png[/img_thumb][/QUOTE]
Nope you are only given what I wrote in the OP. Also I tried your way assuming the ratio's would be the same but as you can see the result for one of them ends up being over 800 and in the problem it says the max distance is 500.
Use the range equation. V^2*sin2(theta)/g Set it equal to 500 which means that that theta is 45 degrees. Solve for V. Plug v and 350 into the equation to get the angle for that. Plug in v and then 14 and 76 degrees to find the distances for those.
[QUOTE=united forever;25671964]Nope you are only given what I wrote in the OP. Also I tried your way assuming the ratio's would be the same but as you can see the result for one of them ends up being over 800 and in the problem it says the max distance is 500.[/QUOTE]
since 76 is bigger than 45, assume 76 is x, the angle is equivalent to |90-x| in degrees.
The bullet's velocity's x-component remains constant and y-component changes because of gravity. The displacement (500m) is caused by the x-component and the change in height (which is 0m) is caused by the y-component [i]and[/i] acceleration.
[img]http://www.codecogs.com/gif.latex?%20\Delta%20x%20=%20v_x%20t[/img]
[img]http://www.codecogs.com/gif.latex?%20\Delta%20h%20=%20v_y%20t%20+%20{1%20\over%202}%20g%20t^2[/img]
Where
[img]http://www.codecogs.com/gif.latex?%20v_x%20=%20v%20\cos(a)[/img]
[img]http://www.codecogs.com/gif.latex?%20v_y%20=%20v%20\sin(a)[/img]
(Draw a triangle of the bullet flying at 45 degrees)
So you have just two unknowns, v and t, and you have two formulas so that's enough to solve both of them. Of those, only v is useful in future.
To write [img]http://www.codecogs.com/gif.latex?\Delta%20x[/img] as a function of the angle, you would solve t from the second equation ([img]http://www.codecogs.com/gif.latex?%20\Delta%20h%20=%20...[/img]) and replace the t in the first equation with that.
at 76 degrees it's 500/45 * 14 = 155.555
Does it give the weight for the bullet? Or does it matter?
[QUOTE=Richard Simmons;25670968]Answer is: Doesn't matter, keep shooting all the kids in the school.[/QUOTE]
[media]http://2.bp.blogspot.com/_CFA81NyTvFo/SjlpgaHdGXI/AAAAAAAAFog/Fx7dSAGZAQo/s400/smiley-get_the_fuck_out.jpg[/media]
Murdering kids at school is not funny
-snip-
[QUOTE=ROFLBURGER;25672045]Does it give the weight for the bullet? Or does it matter?
[/QUOTE]
The weight is negligible, but there is still 9.8 m/s^2 towards the x axis due to gravity
You know the angle of maximum range is 45 degrees without air resistance. Use that to find the velocity. Use that to determine which angle would produce a range of 350m.
Really not difficult.
sorry.
I give up.
:saddowns:
do what ThePuska said.
[QUOTE=ThePuska;25672030]The bullet's velocity's x-component remains constant and y-component changes because of gravity. The displacement (500m) is caused by the x-component and the change in height (which is 0m) is caused by the y-component [i]and[/i] acceleration.
[img_thumb]http://www.codecogs.com/gif.latex?%20\Delta%20x%20=%20v_x%20t[/img_thumb]
[img_thumb]http://www.codecogs.com/gif.latex?%20\Delta%20h%20=%20v_y%20t%20+%20{1%20\over%202}%20g%20t^2[/img_thumb]
Where
[img_thumb]http://www.codecogs.com/gif.latex?%20v_x%20=%20v%20\cos(a)[/img_thumb]
[img_thumb]http://www.codecogs.com/gif.latex?%20v_y%20=%20v%20\sin(a)[/img_thumb]
(Draw a triangle of the bullet flying at 45 degrees)
So you have just two unknowns, v and t, and you have two formulas so that's enough to solve both of them. Of those, only v is useful in future.
To write [img_thumb]http://www.codecogs.com/gif.latex?\Delta%20x[/img_thumb] as a function of the angle, you would solve t from the second equation ([img_thumb]http://www.codecogs.com/gif.latex?%20\Delta%20h%20=%20...[/img_thumb]) and replace the t in the first equation with that.[/QUOTE]
I've found both V and T just in slightly different ways but the reason I came on here is because I got an answer like 50m/s for the velocity, which just seemed way too unrealistic of a result for the velocity of a bullet.
[QUOTE=JohnnyMo1;25672135]You know the angle of maximum range is 45 degrees without air resistance. Use that to find the velocity. Use that to determine which angle would produce a range of 350m.
Really not difficult.[/QUOTE]
Fuck me, forgot about air resistance and the speed of the wind
[QUOTE=ThePuska;25672030]The bullet's velocity's x-component remains constant and y-component changes because of gravity. The displacement (500m) is caused by the x-component and the change in height (which is 0m) is caused by the y-component [i]and[/i] acceleration.
[img_thumb]http://www.codecogs.com/gif.latex?%20\Delta%20x%20=%20v_x%20t[/img_thumb]
[img_thumb]http://www.codecogs.com/gif.latex?%20\Delta%20h%20=%20v_y%20t%20+%20{1%20\over%202}%20g%20t^2[/img_thumb]
Where
[img_thumb]http://www.codecogs.com/gif.latex?%20v_x%20=%20v%20\cos(a)[/img_thumb]
[img_thumb]http://www.codecogs.com/gif.latex?%20v_y%20=%20v%20\sin(a)[/img_thumb]
(Draw a triangle of the bullet flying at 45 degrees)
So you have just two unknowns, v and t, and you have two formulas so that's enough to solve both of them. Of those, only v is useful in future.
To write [img_thumb]http://www.codecogs.com/gif.latex?\Delta%20x[/img_thumb] as a function of the angle, you would solve t from the second equation ([img_thumb]http://www.codecogs.com/gif.latex?%20\Delta%20h%20=%20...[/img_thumb]) and replace the t in the first equation with that.[/QUOTE]
Breaking it into two one-dimensional equations is super unnecessary. You can just use the range equation relaxin posted.
[editline]26th October 2010[/editline]
[QUOTE=ROFLBURGER;25672590]Fuck me, forgot about air resistance and the speed of the wind[/QUOTE]
I'm gonna take a wild guess and say you can neglect those.
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