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A ball is thrown upward near the surface of the earth with a velocity of 50 m/s. After...

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saj-94 | (Level 1) Salutatorian

Posted August 20, 2013 at 3:02 PM via web

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A ball is thrown upward near the surface of the earth with a velocity of 50 m/s. After 5 s later its velocity becomes zero momentarily. If the ball were thrown up with the same velocity on Planet X, after 5 s it would still be moving upwards at 31 m/s. (Neglect air resistance) 
 
The ratio of the magnitude of the gravitational field intensity near the surface of planet X and the magnitude of the gravitational field intensity near the surface of the earth is equal to
 
[1] 0.16  

[2] 0.38  

[3] 0.53  

[4] 0.63  

[5] 1.59

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted August 20, 2013 at 3:24 PM (Answer #1)

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When the ball is thrown upward near the surface of the earth with a velocity of 50 m/s it comes to a halt after 5 seconds. On Planet X, when thrown with the same velocity it continues to move upwards at 31 m/s.

If the initial velocity of a particle is u and its acceleration is a, the final velocity after t seconds is v. The relation between the variables is v = u + a*t.

Let the ratio of the magnitude of the gravitational field intensity near the surface of planet X and the magnitude of the gravitational field intensity near the surface of the Earth be X; if the former is X*C, the latter is X

Using the data given in the problem,

31 = 50 - X*C*5, 0 = 50 - C*5

C*5 = 50

Substitute in 31 = 50 - X*C*5

=> 31 = 50 - 50*X

=> 50*X = 19

=> X = 0.38

The required ratio is 0.38.

The correct answer is option 2.

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