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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11
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Two buildings are 40m apart with what velocity should a ball be thrown horizontally from a window 145 m above the ground in one building so that it will enter a window $22.5 m$ above the ground in second building? $(g=9.8 m/s^2)$

\[(a)\;4 m/s \quad (b)\;6 m/s \quad (c)\;12 m/s \quad (d)\;8 m/s\]

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1 Answer

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Let 'u' be velocity of projection.
't' time taken for ball to fall through a vertical height of $145-22.5=122.5m$
Horizontal distance $=40 m$
$s=ut+\large\frac{1}{2}$$g+2$
$122.5=\large\frac{1}{2} $$ \times 9.8 \times t^2$
(since the vertical component of initial velocity is zero)
$t=5 \;seconds$
Also $u \times t=horizontal \;distance$
$40=u \times 5$
$u=8 m/s$
Hence d is the correct answer. 

 

answered Jun 29, 2013 by meena.p
edited Jan 25, 2014 by meena.p
 

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