An object is thrown with speed 20 m/Sunder the angle pi/3 rad with the horisontal direction. In the highest point, the object is divised into two pieces with same weights. One of them, after the division has the speed 0. How far from the starting point do the pieces fall on ground?
My answer so far is:
The piece that has the speed 0: $$x1=v0^2sin2a/2g=17.32 m$$ $$x2=v0^2sin2a/g=34.64 m$$
The answer in the book is 10, 20 meters.
The horizontal component of velocity of the projectile is $u\cos\theta$ up to the point at which it splits in two, at the highest point. At this point you must apply conservaion of linear momentum. One half immediately stops - its momentum becomes zero - and falls straight down to the ground. The other half continues forward; to conserve momentum it must have twice the velocity of the whole projectile : ie $v=2u\cos\theta$.
From the highest point both halves will take the same time to fall back to the ground, which is the same time it took to reach this point. Because the continuing half has twice the initial horizontal speed, its horizontal distance will be twice the initial value : $x2=2x1$.
Unless I have misunderstood the question, I think the answers given in the book must be wrong. Your answers seem to be correct, if $x2$ is the horizontal distance travelled by the 2nd half of the projectile. But this is not its final distance from the starting point.
