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Physics-

General

Easy

Question

A balloon gong upward with a velocity of $12 blank m s to the power of negative 1 end exponent$ is at a height of 65 m from the earth’s surface at any instant. Exactly at this instant a ball drops from it. How much time will the ball take in reaching the surface of earth?
$open parentheses g equals 10 blank m s to the power of negative 2 end exponent close parentheses$

5 s
6 s
10 s
None of these

The correct answer is: 5 s

From equation of motion, we have
$h equals u t plus fraction numerator 1 over denominator 2 end fraction g t to the power of 2 end exponent$
taking upward direction as negative and downward direction as positive, we have
$blank h equals 65 blank m comma$
$blank u equals negative 12 blank m s to the power of negative 1 end exponent a n d blank g equals 10 blank m s to the power of negative 2 end exponent$
$therefore blank 65 equals negative 12 t plus fraction numerator 1 over denominator 2 end fraction cross times 10 cross times t to the power of 2 end exponent$
$therefore blank 5 t to the power of 2 end exponent minus 12 t minus 65 equals 0$
$rightwards double arrow blank open parentheses t minus 5 close parentheses open parentheses 5 t plus 13 close parentheses equals 0$
$therefore blank t equals 5 blank s$

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