1. ## prove this:projectiles questions

Object is projected from point O passes through point A and B.vertical distance between O-A is am,horizontal distance is bm.Vertical distance to B is bm and horizontal distance is am.

show that object's horizontal range is

(a²+ab+b²)m/(a+b)

what I did was:

O-A
bm=V cos Θ*t
t=x/V cos Θ

O-A
am=v Sin Θ*t-g/2*t²
am=V sin Θ(bm/V cos Θ)-g/2 *(bm/V cosΘ)²
A-B
am=V cos Θ*t
t=x/V cos Θ
A-B
bm=v Sin Θ*t-g/2*t²
bm=V sin Θ(am/V cos Θ)-g/2 *(am/V cosΘ)²
got two equations.but now what??

2. Originally Posted by silvercats
Object is projected from point O passes through point A and B.vertical distance between O-A is am,horizontal distance is bm.Vertical distance to B is bm and horizontal distance is am.

show that object's horizontal range is

(a²+ab+b²)m/(a+b)

what I did was:

O-A
bm=V cos Θ*t
t=x/V cos Θ

O-A
am=v Sin Θ*t-g/2*t²
am=V sin Θ(bm/V cos Θ)-g/2 *(bm/V cosΘ)²
A-B
am=V cos Θ*t
t=x/V cos Θ
A-B
bm=v Sin Θ*t-g/2*t²
bm=V sin Θ(am/V cos Θ)-g/2 *(am/V cosΘ)²
got two equations.but now what??
consider the parabolic trajectory of the projectile in the x-y plane ... it passes thru the points (0,0) , (b,a) , and (a,b)

mathematically, an inverted parabola that passes thru the origin can be written in the form

y = kx(R-x)

where k is a constant and R is the 2nd x-intercept (the "Range" in this case)

for point (b,a) ...

a = kb(R-b)

for point (a,b) ...

b = ka(R-a)

now use the two equations to solve for R in terms of a and b.

3. ah I didn't think we have to use that kind of equations for this.
can't we do this using only kinematics equations?

4. Originally Posted by silvercats
ah I didn't think we have to use that kind of equations for this.
can't we do this using only kinematics equations?
you can, but you'll end up having to eliminate the parameter of time anyway to arrive at the range in terms of a and b ... you end up with essentially the same equation I derived, just a bit messier w/ the constants.