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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Motion in a Straight Line
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Raju drops a pile of marbles from the top of a roof located $8.52\;m$ above the ground. Determine the time required for the marbles to reach the ground .

$\begin{array}{1 1} 1.32\;s \\ 3.5s \\ 2s \\7s\end{array} $

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1 Answer

+1 vote
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 accelerationof the object; the symbol $'v_i'$ stands for the initial velocity value; and the symbol $'v_f'$ f stands for the final velocity $'g'$
An object in free fall experiences an acceleration of -9.8 m/s/s. (The -sign indicates a downward acceleration.) Whether explicitly stated or not, the value of the acceleration in the kinematic equations is $-9.8 \;m/s/s$ for any freely falling object.
$-8.52 \;m =(0 m/s) * (t) +0.5 * (-9.8 m/s2)*(t)*2$
$-8.52 m=(0\;m) *(t)+(-4.9\;m/s2) *(t)2$
$-8.52m=(-4.9 m/s2) *(t)2$
$t=1.32 s$
answered Aug 19, 2014 by meena.p
 

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