Situation 1:
u = v1
v = 0 m/s
a = -g = -9.81m/s^2
S = y1
use S = ut+(1/2)*at^2 to get y1 in terms of v1
Situation 2:
u = 1.27v1
v = 0 m/s
a = -g(saturn) = -0.145m/s^2
S = y2
use S = ut+(1/2)*at^2 to get y2 in terms of v1
Then use the two answers together to get y2 in terms of y1
A 4.03-kg object is thrown vertically upward from the surface of Earth, where the acceleration due to gravity is g1 = 9.81 m/s2. The initial velocity is v1, and the object reaches a maximum height of y1. What is the maximum height, y2, if the object is thrown with a speed of v2 = 1.27v1 from the surface of Saturn\'s moon Tethys? The acceleration due to gravity on Tethys is g2 = 0.145 m/s2. Give your answer as a multiple of y1.
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