To solve this problem, we need to analyze the motion of the projectile before and after it breaks into two equal masses. The key here is to understand the principles of projectile motion and how the break affects the trajectory of the second mass.
Understanding Projectile Motion
When a projectile is launched, it follows a parabolic path due to the influence of gravity. The range of a projectile is determined by its initial velocity, launch angle, and the acceleration due to gravity. In this case, we know the range is supposed to be 1000 meters.
Breaking at the Highest Point
At the highest point of its trajectory, the vertical component of the velocity is zero, while the horizontal component remains unchanged. When the projectile breaks into two equal masses, one mass falls straight down, while the other continues to move horizontally.
Calculating the Horizontal Motion
Let’s denote the time taken to reach the highest point as \( t_{up} \). The total time of flight for the projectile can be calculated as \( T = 2t_{up} \). The horizontal distance covered during this time is equal to the range, which is 1000 meters.
Since the projectile breaks into two parts at the highest point, the mass that continues to move horizontally will maintain its horizontal velocity. The time it takes for this mass to fall to the ground after the break is the same as the time it took to reach the highest point, \( t_{up} \).
Distance Calculation
Let’s denote the horizontal velocity of the projectile as \( v_x \). The horizontal distance traveled by the second mass after the break can be calculated using the formula:
- Distance = Velocity × Time
Since the mass continues to move horizontally for an additional \( t_{up} \), the distance it travels after the break is:
Distance = \( v_x \cdot t_{up} \)
Finding the Values
From the initial conditions, we know that:
- The total range is 1000 meters.
- The horizontal distance covered until the break is 500 meters (since it breaks at the highest point).
Thus, the horizontal velocity \( v_x \) can be expressed as:
Range = \( v_x \cdot T \) = 1000 meters
Since \( T = 2t_{up} \), we can find \( v_x \) and subsequently calculate the distance traveled by the second mass after the break.
Final Distance for the Second Mass
After the break, the second mass will continue to travel horizontally for the same time it took to reach the highest point. Therefore, it will cover an additional 500 meters horizontally after the break, leading to a total distance of:
Total Distance = 500 meters (before break) + 500 meters (after break) = 1000 meters
However, since one mass falls straight down, the second mass will continue to travel horizontally for the same duration it took to reach the highest point, effectively doubling its distance from the launch point. Therefore, the total distance from the launch point becomes:
Distance = 1000 meters (initial range) + 500 meters (additional distance) = 1500 meters.
Conclusion
Thus, the second mass will fall at a distance of 1500 meters from the launching point. The correct answer is (A) 1500 meters.
