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Q.

A boat in still water is moving directly away from a cannon on the shore, with a speed v1. The cannon fires a shell with a speed v2  at an angle α  with horizontal and the shell hits the boat. Then The distance of the boat from the cannon at the instant the shell is fired is  xg(v2sinα)(v2cosα−v1), then x =

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answer is 2.

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Detailed Solution

Time of flight  =2v2sinαgDistance travelled by the boat  =2v1v2sinαgRange =  v2cos⁡α2v2sin⁡αg∴  Distance of the boat from the cannon at the instant the shell is fired is v2cos⁡α2v2sin⁡αg−v12v2sin⁡αg =2g(v2sinα)(v2cosα−v1)
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