While attempting a landing on the moon, astronauts had to change their landing site and land at a spot that was 4 kilometers away from the original site. Assuming that they were at a height of 137 meters, calculate the horizontal velocity of the spacecraft during touchdown if it lands in a free-fall mode without using retro engines. Consider gravity = 1.63 meters/second2.
A = original landing site. B = alternate landing site.
Assume that aerodynamic resistance is negligible.
The craft lands in free-fall mode, therefore its original vertical velocity is zero. It travels downward by 137 m at gravitational acceleration of g = 1.63 m/s². The time of travel, t, obeys the equation (137 m) = (1/2)(1.63 m/s²)*(t s)² Therefore t² = (137*2)/1.63 = 168.098 => t = 12.965 s.
The constant horizontal velocity, u, required to travel 4 km (or 4000 m) in time, t, is given by (u m/s)*(12.965 s) = (4000 m) u = 4000/12.965 = 308.5 m/s (nearest tenth)
