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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Motion in a Plane
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A body is thrown horizontally from top of a tower and strikes the ground after three seconds at an angle $45^{\circ}$ with horizontal. Find the height of the tower and initial speed of body.

\[(a)\;98.2\; m,12.2\; m/s \quad (b)\;19.8\; m,14.6 m/s \quad (c)\;12.1\;m,19.2 m/s \quad (d)\;44.1\;m,29.4 m/s\]

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For the body thrown horizontally
Therefore $s_y=u_yt+\large\frac{1}{2} $$\times 9.8 \times 3^2$
$=44.1 m$
[Let $v_v$ and $v_x$ be the vertical & horizotal component when it strikes the ground]
$ \qquad= 0+9.8 \times 3$
$ \qquad= 29.4 m/s$
The body strikes the ground at $45^{\circ}$
$\tan 45=\large\frac{v_y}{v_x}=>\frac{29.4}{v_x}$$=1$
Therefore $v_x=29.4$
Therefore the horizontal speed at the time of projection is same as horizontal speed at the time of striking the ground
Therefore $u=29.4$
Hence d is the correct answer.


answered Jul 2, 2013 by meena.p
edited Jan 26, 2014 by meena.p

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