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

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For the body thrown horizontally

$u_x=u$

$u_y=0$

$a_y=9.8$

$t=3$

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.

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