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From the top of a tower, a ball is thrown vertically upward whcih reaches the ground in \mathrm{6} s.
A second ball thrown vertically downward from the same position with the same speed reaches the ground in \mathrm{1.5\, s}. A third ball released, from the rest from the same location,will reach the ground in _____s.

Option: 1

3


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

                                
We know that ,
\mathrm{t_{1}= \sqrt{t_{2}t_{3}}}
\mathrm{t_{1}= \sqrt{9}= 3s}

Ball released from the rest from the same location will reach ground after \mathrm{t_{1}= 3s}
 

Posted by

Nehul

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