Question

a particle is projected such taht it just passes the 4 corners of a regular hexagon ...find its range???

Answer 1

it will not be a triangle ...instead a parabola ....it is a case of projectile!!

Answer 2

parabola .... ok; but the set up is good otherwise?

we start at one end of the 'range' and shoot it over the hex in this manner and it lands on the other side?

Answer 3

you got it...correct

Answer 4

the only other info I would need is; what length are the sides of a regular hexagon? or are we to keep this thing generalized?

Answer 5

all sides are equal to ... 'a'

Answer 6

lol .... then generalized it is :)

Answer 7

yea...i mean the same thing

Answer 8

let me work out some drawings and get my concepts in order; about 15 minutes tops ...

Answer 9

go ahead!

Answer 10

I think it would be 3a

Answer 11

I know that you only need 3 points to establish a parabola with; so im gonna take these 3 points and do some 'specific' tests for a; such as a=1 and a=5, then I can test to see if the range is a scalar or not :)

Answer 12

and the top there should be a, 2a sin(60) :)

Answer 13

answer is

Answer 14

a\[\sqrt{7}\]

Answer 15

answer is a=sqrt(7)? or range = sqrt(7) ?

Answer 16

range is ...... a.sqrt(7)

Answer 17

ahh ... that makes more sense :) how did you determine that?

Answer 18

when I make a = 1, I get roots of the parabola that are 2.645751311..... apart; which is the sqrt(7)

Answer 19

If we place the origin at the center of the hexagon, then the ground will be line
y = -a*sqrt(3)/2. We can work with the vertices at (a,0) and (a/2, a*sqrt(3)/2) to determine the parabola passing through those two points. The parabola is symmetric about the y-axis and is concave down, so its equation is
y= u - v*x^2. We need to determine what u and v are so that the parabola passes through those points. For (a,0) we have 0 = u - a^2*v, and for (a/2, a*sqrt(3)/2), we have a*sqrt(3)/2 = u - (3*a^2/2)*v. That gives us u = 2a*sqrt(3) and
v = 2*sqrt(3)/a. Therefore, the equation of the trajectory is
y = 2a*sqrt*3) - (2*sqrt(3)/a)*x^2. Now we need to determine where this parabola intersects the line y = -a*sqrt(3)/2 and get the distance between those two points.
So we now have 2a*sqrt*3) - (2*sqrt(3)/a)*x^2 = -a*sqrt(3)/2, making
x = +-a*sqrt(6)/2. The distance between those points of intersection, and hence the range, is a*sqrt(6).