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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Motion in a Plane
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A ball slides off a horizontal table top $1.25 \;m$ hight, with a velocity of $4ms^{-1}$ Find the horizontal distance from the edge of the table at which the ball strikes the ground. $(g= 10 ms^{-2})$

$\begin{array}{1 1}(A)\;100\;m \\(B)\;2.0\;m \\(C)\;300\;m\\(D)\;6m \end{array} $

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The motion of ball is composed of two motions (i) horizontal motion with velocity $4 ms^{-1} $
(ii) vertical motion with $u=0;a=10ms^{-2},s= 1.25 m$.
If the times to strike the ground is t seconds, then $1.25 =\large\frac{1}{2} $$(10)t^2, t= 0.5 s$
Horizontal distance $= v_0 t = 4 \times 0.5 =2.0 m$
Hence B is the correct answer.
answered Jul 16, 2014 by meena.p
 

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