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

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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.

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