At an elevation of 56.7 km above the surface of mars, your spacecraft is dropping vertically at a speed of 265 m/s. the spacecraft is to make a soft landing -- that is, at the instant it reaches the surface of mars, its velocity is zero. assume the spacecraft undergoes constant acceleration from the elevation of 56.7 km until it reaches the surface of mars. what is the magnitude of the acceleration?

0.62 m/s²

Explanation:

Consider the motion of spacecraft in vertical direction

= vertical displacement of the spacecraft = elevation oabove the surface of mars = 56.7 km = 56700 m

= initial speed of spacecraft at the given elevation = 265 m/s

= final speed of spacecraft on the surface of mars = 0 m/s

a = acceleration experienced by the spacecraft

Using the kinematics equation based on the data available, we have

hence the magnitude of acceleration is 0.62

jupiter is the largest planet in our solar system, having a mass and radius that are respectively, 318 and 11.2 times that of earth. suppose that an object falls from rest near the surface of each planet and that the acceleration due to gravity remains constant during the fall. each object falls the same distance before striking the ground. determine the ratio of the time of fall on jupiter to that on earth.