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A ball is thrown from horizontal level. Such that it clears a wall 6 m high, situated at horizontal distance of 35 m. If the angle of projection is 600 w.r.t. horizontal,

what should be the minimum velocity of projection?

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Let minimum velocity of projection be v

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$s=35\\ U=v\cos 60 \\ a=0 \\ \text {using } ,s=vt \\ 35= v.\cos60*(t) \\ t=35/(v\cos 60)----(1) \\ \text { vertical motion } \\ S=6 \\ u=v\sin 60 \\ a=-9.81 \\ \text {using } s=ut+(1/2)at^2 \\ 6=v\sin 60(t) +(1/2)(-9.81)t^2 $

Putting value of t from eqn (1)

$6=v\sin 60(35/v\cos 60) - 0.5 (9.81)(35/v\cos 60)^2 \\ 6=35 (\tan 60) -4.905(35)^2/ (v^2\cos^2 60) \\ 6=60.62 -24034.5 /v^2 \\ \text {solving, we get } \\ V=20.98 m/s$

Minimum velocity of projection is v=20.98 m/s

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