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=

http://web.mit.edu/8.01t/www/materials/InClass/IC_Sol_W05D1-2.pdf

To solve this problem, we need to apply principles of projectile motion.

Let's start by analyzing the motion of the smaller piece, m2. Since it returns to the launching station, we know that its horizontal displacement, xf, must be equal to xm.

We can use the equation for horizontal displacement in projectile motion, which is given by:

xf = v_i * t

where v_i is the initial horizontal velocity of the object and t is the time of flight.

Since the smaller piece m2 lands at the same height it was launched from, the time of flight for m2 can be determined by using the equation for vertical displacement in projectile motion, which is given by:

yf = v_iy * t + (1/2) * g * t^2

where v_iy is the initial vertical velocity (which is equal to 0 when the object is launched horizontally) and g is the acceleration due to gravity.

Since yf is equal to 0 (the object returns to the same height it was launched from), we can rearrange the equation to solve for t:

0 = (1/2) * g * t^2

Simplifying, we get:

t = sqrt(0) = 0

This means that the time of flight for m2 is 0, indicating that it returns to the launching station immediately.

Therefore, the horizontal displacement, xf, is equal to xm (which is the distance between the launch point and the explosion).

xf = xm

Now let's analyze the motion of the larger piece, m3. Since it has three times the mass of m2, we expect it to have the same horizontal displacement, xf = xm.

Using the same equation for horizontal displacement, we can solve for the initial horizontal velocity, v_i3, of m3:

xf = v_i3 * t

Since xf = xm and t = 0 for m2, we can substitute those values in the equation:

xm = v_i3 * 0

Since t = 0, the initial horizontal velocity, v_i3, for m3 is also 0.

Therefore, the larger piece m3 falls straight down from the point of explosion, landing at the same location as the smaller piece m2.

Hence, xf = xm