A fireworks projectile is launched upward at an angle above a large flat plane. When the projectile reaches the top of its flight, at a height of \(h\) above a point that is a horizontal distance \(D\) from where it was launched, the projectile explodes into two equal pieces. One piece reverses its velocity and travels directly back to the launch point. How far from the launch point does the other piece land? a) \(D\) b) \(2 D\) c) \(3 D\) d) \(4 D\)

First, let's apply the conservation of momentum for the horizontal motion. Since the total momentum before and after the explosion remains constant, we can find the velocity of the second piece after the midpoint. Let \(v_1\) be the horizontal velocity of the first piece after the explosion, and let \(v_2\) be the horizontal velocity of the second piece. Since both the masses are equal, we can write the conservation of momentum equation as: \(m v_1 = m v_2\) where \(m\) is the mass of the individual pieces. This gives us the relation: \(v_1 = v_2\).

At the highest point, the first piece reverses its direction and travels the same horizontal distance back to the launch point. That means, the horizontal component of velocity for the first piece remains the same (but in the opposite direction) throughout its motion. This implies that the horizontal velocity of the first piece at the highest point (\(v_1\)) is equal to the horizontal velocity of the entire projectile at that point. We can call the horizontal initial velocity of the projectile \(v_{0x}\) (which is equal to \(v_1\)).

Using the conservation of momentum relation derived in step 1, we can find the horizontal velocity of the second piece: \(v_2 = v_1\) From step 2, we know \(v_1 = v_{0x}\). So, \(v_2 = v_{0x}\)

Since the second piece has the same horizontal velocity as the initial projectile velocity, it will travel horizontally while falling from a height \(h\) to the ground. Using the vertical component of motion, we can find the time of flight for this piece. The vertical motion equation is: \(h = \frac{1}{2} g t^2\) where \(g\) is the acceleration due to gravity, and \(t\) is the time of flight. So, \(t = \sqrt{\frac{2h}{g}}\)

Now, we can find the distance traveled by the second piece using its horizontal velocity \(v_2\) and the time of flight \(t\): Horizontal distance \(= v_{2}t\) Substituting the values of \(v_2\) and \(t\), Horizontal distance \(= v_{0x} \sqrt{\frac{2h}{g}}\) Since the first piece travels a horizontal distance \(D\) before reaching the half-point, the total horizontal distance traveled by the projectile at this point can be expressed as: Total distance \(= D + v_{0x} \sqrt{\frac{2h}{g}}\) As the answer format is D, we can make the equation have a factor of D: Total distance \(= D(1 + \frac{v_{0x}}{D} \sqrt{\frac{2h}{g}})\) The coefficient of D in this equation represents the answer to the problem, which can be compared to the given options: After comparing the options, we can conclude that the other piece lands at a distance of: Total distance \(= 2D\). So, the correct answer is (b) \(2 D\).