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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Motion in a Straight Line
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From the top of a building 40 m tall, a ball a thrown vertically upwards with a velocity of $10\; ms^{-1}$ After how long will the ball hit the ground?

$\begin{array}{1 1} 4s \\ 3s \\ 2s \\7s\end{array} $

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$s= -40 m, u= +10 ms^{-1} ,a=-10 ms^{-2}$
Now $s= ut+ \large\frac{1}{2}$$ at^2$
=> $ -40 -10t+\large\frac{1}{2} $$\times (-10) t^2$
=> $r^2-2t -8 =0$
=> $(t+2) (t-4) =0$
=> $t=-2 \;or\; 4 s$
The negative value of t is not possible
Hence the ball will hit the ground after 4 s
answered Aug 19, 2014 by meena.p

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