An instrument-carrying projectile of mass m1 accidentally explodes at the top of its trajectory. The horizontal distance between launch point and the explosion is xm. The projectile breaks into two pieces which fly apart horizontally. The larger piece, m3, has three times the mass of the smaller piece, m2. To the surprise of the scientist in charge, the smaller piece returns to earth at the launching station. Neglect air resistance and effects due to the earth's curvature.

How far away, xf, from the original launching point does the larger piece land? Express your answer in terms of some or all of the given variables m1, xm, and g (enter m_1 for m1 and x_m for xm).

xf=

To find the horizontal distance xf where the larger piece lands, we can use the principle of conservation of momentum in the horizontal direction.

Let's denote the velocity of the smaller piece m2 as v2 and the velocity of the larger piece m3 as v3.

Since the smaller piece m2 returns to earth at the launching station, we know its final velocity, vf2, is 0.

Using the principle of conservation of momentum, we have:
(m2 * v2) + (m3 * v3) = 0

Since m3 has three times the mass of m2, we can rewrite this as:
(m2 * v2) + (3 * m2 * v3) = 0

Dividing through by m2:
v2 + 3v3 = 0

Now let's consider the initial velocities of the two pieces. Since the explosion happened at the top of the projectile's trajectory, both pieces were initially moving only in the horizontal direction. Therefore, their initial vertical velocities can be ignored.

The only horizontal velocity component is the initial velocity of the projectile, which we'll call vxi.

For the smaller piece m2, we have:
v2 = vxi

For the larger piece m3, we have:
v3 = vxi

Substituting v2 and v3 back into the equation v2 + 3v3 = 0:
vxi + 3vxi = 0
4vxi = 0
vxi = 0

This means that both the smaller piece m2 and the larger piece m3 had an initial velocity of 0 in the horizontal direction, as there is no initial horizontal velocity component.

Therefore, the larger piece m3 does not move horizontally and lands at the location of the explosion. The horizontal distance xf from the original launching point is equal to the horizontal distance between the launch point and the explosion, xm.

So, the answer is:
xf = xm