The Impact of External Factors on Projectile Range

[perfectz]

Projectile doubt...

CHOOSE THE CORRECT ANSWER AND EXPLAIN.

What is the nature of proportionality between the muzzle angle and the horizontal distance traveled by a projectile?

a) Directly proportional.
b) Inversely proportional.
c) None of the above.


( who says learning physics is boring? )

 

[Doc Al]

Well, what do you think?

 

[husky88]

Well think about this. The bigger the angle, the higher the projectile goes. The higher it goes, the longer time it takes to reach the ground. The longer it takes to reach the ground,...
You don't even need formulas for this.

 

[Doc Al]

The bigger the angle, the higher the projectile goes.

True.

The higher it goes, the longer time it takes to reach the ground.

Also true.

The longer it takes to reach the ground,...

Not sure what you can immediately deduce from this.

You don't even need formulas for this.

You might want to reconsider that.

 

[husky88]

Well, what I thought was the longer it takes to reach the ground, the more time it has to travel horizontally, therefore the more horizontal distance it travels. Skipping over, you get:
The bigger the angle, the more horizontal distance it travels.
So they are directly proportional.

 

[Weimin]

So what about the angle of 90 degrees? :-)

 

[husky88]

Hmm. It does say it is a muzzle angle, so if the angle is 90, then you end up in shooting yourself and then you wouldn't care about the answer anyway. :)

Yeah, I guess my non-formula logic doesn't apply for a 90 angle. Then the answer would be C.

 

[husky88]

I just realized that if you throw it too high, then Vx will decrease.
So yeah, all my posts don't make sense. Except for the one with the 90 degree angle. :)

Could you say it is proportional to sin(2*angle), then?

 

[cristo]

Could you say it is proportional to sin(2*angle), then?

What makes you think that?

 

[perfectz]

Am I right?

i too came to Husky's answer before posting the topic. But i had doubts whether it was right or wrong.

let the angle be 'A'.
Let the velocity be 'V'
therefore
horizontal component = V cos A
vertical component = V sin A

therefore
horizontal displacement = V cos A * t(time of flight) units
vertical displacement = V sin A * t units

At maximum height,
V sin A = 0. ----------------> l

consider formula V = U-gt ----------------> ll


therefore substituting l in ll

0 = V sin A - gt
t = (V sin A)/g

total time of flight = 2t
=2((V sin A)/g)

range is nothing but
the horizontal displacement V cos A * t

so range = (2(V sin A)/g} * V cos A

therefore range is directly proportional to Sin2A

Am I right guys?

 

[perfectz]

dont just visit people
please comment
reply pleaseeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeee

 

[Astronuc]

Correct!

http://hyperphysics.phy-astr.gsu.edu/hbase/traj.html#tra4

So what can one say about R $\varpropto$ sin(2A) versus R $\varpropto$ kA (directly proportional) or $\varpropto$ k/A (inversely proportional)?

 

[perfectz]

ya hooooooooooooooooooooo
Thanks guys.
Physics Forums Rok and you guys are cool
the site u gave me is just beyond cool dude How much does wind affect the range of a projectile body?
And how much does the shape and weight of the body affect the range?