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in Gravitation by (70.8k points)
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A ball which is thrown upward near the surface of the Earth with a velocity of 50 m/s will come to rest about 5 seconds later. If the ball were thrown up with the same velocity on Planet X, after 5 seconds it would be still moving upwards at nearly 31m/s. The magnitude of the gravitational field near the surface of Planet X is what fraction of the gravitational field near the surface of the Earth? 

(A) 0.16 

(B) 0.39 

(C) 0.53 

(D) 0.63 

(E) 1.59

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2 Answers

+1 vote
by (65.2k points)

Correct option (B) 0.39

g = ∆v/t = (31m/s – 50m/s)/(5s) = –3.8m/s2

0 votes
by (15 points)
g on earth is 9.8m/sec square..g on planet X will be find by using equation of motion vf=vi+gt. For moving upward 31=50-g×5

31-50=-g×5

19=g5 ..g=3.8..g on planet/g on earth..3.8/9.8=0.387 approx 0.39...

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