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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Motion in a Straight Line
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A ball is thrown vertically upwards from the top of a tower of height h with velocity v. The ball strikes the ground after

$\begin{array}{1 1} \large\frac{v}{g} \normalsize \bigg[ 1+ \sqrt {1+ \frac{2gh}{v^2}}\bigg] \\ \large\frac{v}{g} \normalsize \bigg[ 1- \sqrt {1+ \frac{2gh}{v^2}}\bigg]\\\large\frac{v}{g} \normalsize \bigg[ 1+ \sqrt {1+ \frac{2gh}{v^2}}\bigg]^{1/2}\\\large\frac{v}{g} \normalsize \bigg[ 1- \frac{2gh}{v^2}\bigg]^{1/2}\end{array} $

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Answer : $ \large\frac{v}{g} \normalsize \bigg[ 1+ \sqrt {1+ \frac{2gh}{v^2}}\bigg]$
$h= -vt +\large\frac{1}{2}$$gt^2$
or $ gt^2 -2vt-h=0$
$t= \large\frac{-(-2v) \pm \sqrt { 4v^2+4gh}}{2g}$
$\quad= \large\frac{2v \pm 2 \sqrt{ v^2+gh}}{2g}$
$\quad= \large\frac{v}{g} \pm \large\frac{[v^2+2gh]^{1/2}}{g}$
$\quad= \large\frac{v}{g} \normalsize \bigg[ 1+ \sqrt {1+ \frac{2gh}{v^2}}\bigg]$
Now retain only the positive sign.
answered Aug 14, 2014 by meena.p

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