Unformatted text preview: positions symmetric about the original target point. Since one lands at , 2 1 R the other lands at m. 848 120 sin ) s / m 80 . 9 ( ) s / m 80 ( 2 3 2 sin 2 3 2 3 2 2 2 = = α = ο g v R b) In terms of the mass m of the original fragment and the speed v before the explosion, 2 1 1 = K . so , ) 2 ( and 2 2 1 2 2 1 2 2 2 2 2 1 2 2 mv mv mv K mv v K mv m == ∆ = = The speed v is related to , cos by α = v v v so J. 10 60 . 1 ) . 60 cos ) m/s 80 )( kg . 20 ( 2 1 cos 2 1 4 2 2 2 × = ° = = ∆ α mv K...
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 Mass, original parabolic trajectory, original target point

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