If a superlaser beam is shot 100km away from the target and installed flat on the ground to a this target on top of a building, how tall should the building so the laser hits the target 100km away?

To solve this problem, we can use a simple geometric approach. Given that the laser is shot from a point at ground level to a target on a building 100 km away, we need to determine the height of the building such that the laser beam hits the target. Assuming the laser beam travels in a straight line, we can use a right triangle to represent the situation: 1. The horizontal distance from the laser to the target is \( 100 \) km (which is \( 100,000 \) meters). 2. Let \( h \) be the height of the building (in meters). 3. The length of the laser beam forms the hypotenuse of the triangle, but since we're working with a flat ground, we can ignore the hypotenuse in this particular case, because as long as the beam is aimed correctly, it will reach the height it needs at the building. For this situation, the conclusion we can draw is that the building must be at least a small height above the ground level to ensure the laser can reach it without any obstruction depending on the curvature of the Earth. ### Curvature of the Earth For an accurate calculation, we can also consider the curvature of the Earth, since the over-distance is considerable. The formula to account for the Earth’s curvature is given by: \[ h = \frac{d^2}{2R} \] Where: - \( h \) is the height of the building, - \( d \) is the distance to the target, - \( R \) is the radius of the Earth (approximately \( 6,371,000 \) meters). Plugging in the distance \( d = 100,000 \) meters: \[ h = \frac{(100,000)^2}{2 \times 6,371,000} \] \[ h = \frac{10,000,000,000}{12,742,000} \] \[ h \approx 785.4 \text{ meters} \] ### Conclusion To ensure that the laser beam, fired from ground level, will correctly hit the target at the top of a building located 100 km away, accommodating for the curvature of the Earth, the building would need to be approximately **785.4 meters tall**.