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Home  >>  JEEMAIN and NEET  >>  Physics  >>  Class11  >>  Motion in a Straight Line

A feather is dropped on the moon from a height of $1.40\; meters$. The acceleration of gravity on the moon is $1.67 m/s^2$. Determine the time for the feather to fall to the surface of the moon

$\begin{array}{1 1} -1.67\;s \\ 1.29 \\ 2.7\;s \\3.6\;s\end{array} $

1 Answer

Given $v_i =o m/s$
$d= -1.40\;m$
$a= -1.67 \;m/s^2$
Use : $d=v_i* t +\large\frac{1}{2}$$*a*t^2$
$d$ stands for the displacement;
the symbol $t$ stands for the time;
the symbol $'a'$ stands for the acceleration of the object;
the symbol 'v_i$ stands for the initial velocity value;
and the symbol $v_f$ stands for the final velocity.
$d=v_i*t +0.5 *t*t^2-1.40m$
$\quad= (0 m/s) * (t) +0.5 * (-1.67 m/s^2) * (t)^2$
$-1.40 m= 0+ (-0.835 m/s^2)*(t)^2$
$\large\frac{(-1.40 m)}{-0.835 m/s^2)}$$=t^2$
$1.68 s^2 =t^2$
$t= 1.29\;s$
answered Aug 20, 2014 by meena.p

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