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Sagarika
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An acrobat is launched from a cannon at an angle of 60 degrees above the horizontal. The acrobat is caught by a safety net
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Here,velocity of projection,u=30 m/s Angle of projection=30° We have,maximum height attained by a projectile is H=u^2 sin tita/2g (u square sin tita divided by 2g) =30^2×sin 30/2×9.8 =900×1/2/(whole divided by)19.6 =450/19.6 =22.95 =23 m (approx)
