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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.

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

For the body thrown horizontally
$u_x=u$
$u_y=0$
$a_y=9.8$
$t=3$
Therefore $s_y=u_yt+\large\frac{1}{2} $$\times 9.8 \times 3^2 =44.1 m [Let v_v and v_x be the vertical & horizotal component when it strikes the ground] v_y=u_y+a_yt \qquad= 0+9.8 \times 3 \qquad= 29.4 m/s The body strikes the ground at 45^{\circ} \tan 45=\large\frac{v_y}{v_x}=>\frac{29.4}{v_x}$$=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.

edited Jan 26, 2014 by meena.p

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