A 3.31-kg object is thrown vertically upward from the surface of Venus, where th

Question

A 3.31-kg object is thrown vertically upward from the surface of Venus, where the acceleration due to gravity is g1 = 8.87 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 = 1.15v1 from the surface of Jupiter's moon Ganymede? The acceleration due to gravity on Ganymede is g2 = 1.43 m/s2. Give your answer as a multiple of y1.A 3.31-kg object is thrown vertically upward from the surface of Venus, where the acceleration due to gravity is g1 = 8.87 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 = 1.15v1 from the surface of Jupiter's moon Ganymede? The acceleration due to gravity on Ganymede is g2 = 1.43 m/s2. Give your answer as a multiple of y1.

Explanation / Answer

t1 = v1/ g1

Now, y1 = v1t1 + 0.5 x g1 x t12
y1 = v12/ g1 + 0.5 v12/ g1 = 1.5 v12 / g1----------------- 1

t2 = v2/ g2 = 1.15 v1 / g2

Now, y2 = v2t2 + 0.5 x g2 x t22
y2 = 1.152 v12 / g2 + 0.5 x 1.152 v12 / g2 = 1.5 x 1.152 v12 / g2 -----------------------2

Dividing 1 by 2 :

y1/ y2 = g2/ (g1 x 1.152)

y2 = y1 x g1 x 1.152 / g2 = 8.87 x 1.152 x y1 / 1.43 = 8.2 y1

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