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 = v2cosα2v2sinαg∴ Distance of the boat from the cannon at the instant the shell is fired is v2cosα2v2sinαg−v12v2sinαg =2g(v2sinα)(v2cosα−v1)