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A spaceship hovering over the surface of Uranus drops an object from a height of 19 m. How much longer does it take to reach the surface than if dropped from the same height on Earth? Neglect air resistance in both cases. [The acceleration due to gravity on Uranus is 88.9% of that on Earth, gUranus = (0.889)g.] answer in : s

Respuesta :

h= height = 19 m

g.e = gravity on earth = 9.8 m/s^2

g.u = gravity on Uranus = 9.8 x (88.9/100) = 8.71 m/s^2

Apply :

h = vot + 1/2 gt^2

h= 1/2 g t^2

t = √(2h/g)

Earth:

t= √(2(19)/9.8) = 1.97 s

Uranus

t= √(2(19)/8.71) = 2.09 s

Tu - Te = 2.09 - 1.97 = 0.12 s

Answer : 0.12 s