physics question (1 Viewer)

[dlesmond]

Hi all I have a difficult projectile question...Well i think so anyway...

A gun has a muzzle velocity of 100m/s. What two angles of elevation could be used to hit a target 300m away?

I have found the all the velocities etc. both horizontally and vertically but don't know how to find these angles?

Any help would be great thanks....

 

[vulgarfraction]

Say an angle is x

range =t 100 cos x

where t 100 sin x - (1/2)gt^2=0 ==> t=0 or 200 sin x/g (start and end)

so the range is (20000 sin x cos x)/g = 10000 sin (2x)/g.

Sub in range = 300, find the possible values for x (between 0 and 90 inclusive). There are two values.

Is this the right forum for this?

 

[SoulSearcher]

Well that is the extension 1 way of getting the answer, I think he wants the physics method, which I don't know yet.

 

[airie]

Well that is the extension 1 way of getting the answer, I think he wants the physics method, which I don't know yet.

That's actually the method we're taught to do it, in physics o.0

 

[~shinigami~]

So have anyone got an answer yet? Cause I keep getting the wrong answer.

 

[airie]

So have anyone got an answer yet? Cause I keep getting the wrong answer.

The method is already given by vulgarfraction above, so what's the problem?

Assuming the target is level with the point of launch, and that g = 9.8ms^-2 as vulgarfraction did, and follow through her method, you get sin 2x = 0.294, so 2x is approx. 17 or 163 degrees (to the nearest degree). Therefore, x is either 9 or 81 degrees (to nearest degree).

 

[SoulSearcher]

That's actually the method we're taught to do it, in physics o.0

Well excuse me for not knowing that yet

 

[~shinigami~]

The method is already given by vulgarfraction above, so what's the problem?

Assuming the target is level with the point of launch, and that g = 9.8ms^-2 as vulgarfraction did, and follow through her method, you get sin 2x = 0.294, so 2x is approx. 17 or 163 degrees (to the nearest degree). Therefore, x is either 9 or 81 degrees (to nearest degree).

Oh shit, I know what I did wrong now.

Instead of dividing 17 and 163 degrees by 2, I multiplied them. What a stupid mistake by me. :rofl:

 

[airie]

Oh shit, I know what I did wrong now.

Instead of dividing 17 and 163 degrees by 2, I multiplied them. What a stupid mistake by me. :rofl:

lol dw, at least you know you wouldn't do that next time

 

[~shinigami~]

lol dw, at least you know you wouldn't do that next time

Thanks airie.

That's the good thing about making really stupid mistakes, I'll never repeat it again.

 

[SoulSearcher]

That's the good thing about making really stupid mistakes, I'll never repeat it again.

I tend to repeat them a lot

 

[m0ofin]

I usually forget what my mistakes are so they're done over and over again, lol, bad habit kids.

 

[~shinigami~]

SS: As if, everyone knows your a math and Phys freak.

Of course, I mean "freak" in a good way though.

 

[dlesmond]

Hey everyone thanks for your help, I finally figured out how to do it!
But it is a different way to the method above......

Firstly I found both of the vertical and horizontal velocties and used a vector diagram to find the angled velocity (using pythagoras). I then plugged it into the equation....

VH = VcosΘ
VV = VsinΘ


Thus leaving two angles, each being 8.36° and -8.36°, the second angle (the minus) is subtracted from 90° giving 81.64°

Thanks again