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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Motion in a Straight Line
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An astronaut standing on a platform on the Moon drops a hammer. If the hammer falls $6.0 $ meters vertically in $2.7 \;seconds$, what is its acceleration ?

$\begin{array}{1 1} 1.6 \;m/s^2 \\ 4.4 \;m/s^2 \\ 2.2\;m/s^2 \\9.8\;m/s^2 \end{array} $

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Given $v_i =0$
$d= 6 \;m$
Find $a=$
$d= v_i t+\large\frac{1}{2}$$at^2$
$a= \large\frac{2d}{t^2}$
$a= \large\frac{2(6m)}{(2.7 s)^2}$
$a= 1.6 m/s^2$
answered Aug 14, 2014 by meena.p

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