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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Motion in a Straight Line
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A particle is projected vertically from a point A on ground. It takes time $t_1$ to reach point B but still continous to move up. It takes further $t_2$ time to reach the ground from point B. Height of point B from the ground is

\[(a)\;\frac{1}{2} g(t_1+t_2)^2\quad (b)\;g t_1t_2 \quad (c)\;\frac{1}{8} g(t_1+t_2)^2 \quad (d)\frac{1}{2} g t_1t_2\]

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1 Answer

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Time taken for the particle to reach highest point is $ \large\frac{t_1+t_2}{2}$
$(v=u-gt)$
Therefore initial velocity$=u=g \bigg(\frac{t_1+t_2}{2}\bigg)$
Therefore heigh of B from ground is
$h=ut+\large\frac{1}{2}$$at ^2$
$h=ut_1-\large\frac{1}{2}$$g{t_1}^2$
$\quad=g \bigg(\large\frac{t_1+t_2}{2}\bigg)$$t_1-\large\frac{1}{2} $$gt_1^2$
$h=\large\frac{1}{2}$$g t_1 t_2$
Hence d is the correct answer.

 

answered Jun 26, 2013 by meena.p
edited Jan 25, 2014 by meena.p
 

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