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There are two values of time for which a projectile is at same height. The sum of these two times is

\[(a)\;\frac{3T}{2}\quad (b)\;\frac{4T}{3} \quad (c)\;\frac{3T}{4} \quad (d)\;T\]

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Time taken to go from $O \;to\; A=t_1$
Time taken to go from $O \;to\; B=t_2$
Also we have time taken to go from $O \;to\; A$=Time taken to go from $B \;to\; D$
Therefore $ t_1+t_2=t_2$+ time taken to go from $B \;to\; D$
Hence d is the correct answer.


answered Jul 1, 2013 by meena.p
edited Jan 26, 2014 by meena.p

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