John and mary are skating at an ice rink. john skates at a constant speed of 6.7 m/s, with respect to the ice surface, directly south. mary skates at a constant speed of 10.9 m/s in a direction of 28â° west of south, with respect to the ice. what are the magnitude and direction of mary's velocity with respect to john? give the direction as measured relative to due south.

60.3° from due south and 5.89 m/s    For this problem, first calculate a translation that will put John's destination directly on the origin and apply that translation to Mary's destination. Then the vector from the origin to Mary's new destination will be the relative vector of Mary as compared to John. So John is traveling due south at 6.7 m/s. After 1 second, he will be at coordinates (0,-6.7). The translation will be (0,6.7)  Mary is traveling 28° West of due south. So her location after 1 second will be  (-sin(28)*10.9, -cos(28)*10.9) = (-5.117240034, -9.624128762)  After translating that coordinate up by 6.7, you get  (-5.117240034, -2.924128762)  The tangent of the angle will be 2.924128762/5.117240034 = 0.57142693  The arc tangent is atan(0.57142693) = 29.74481039° Subtract that value from 90 since you want the complement of the angle which is now 60.25518961°    So Mary is traveling 60.3° relative to due south as seen from John's point of view.  The magnitude of her relative speed is  sqrt(-5.117240034^2 + -2.924128762^2) = 5.893783 m/s    Rounding the results to 3 significant digits results in 60.3° and 5.89 m/s

Wavelength divides in half and speed stays the same.

The objects do not reach terminal velocity         there is not enough time from the time the object falls from when it hits the ground to achieve terminal velocity. an object must be placed higher to achieve terminal velocity in order to be in free fall.