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Shell is fired from a point O at an angle of 60 degree with the horizontal with speed of 40 metres per second .it strikes the horizontal plane through O at a point A. the gun is fired for second time with the same angle of elevation as before but a different speed V. the target starts to rise vertically upwards from A with a constant speed 9 root 3 metre per second at the same instant as the shell is fired if the shelk strikes the target find V
For the first shot the horizontal distance covered is R = 40^2sin120`÷10 = 80root3 m. For the second shot, let the velocity of projection be V, and it hits the target after t seconds. Then vertical displacement of the shot in time t = veritical displacement of the tarfet in time t. I.e, 9root3.t = Vsin60.t - gt^2÷2 => t = (V-18)root3÷10.And the horizontal displacement of the first shot = the horizontal displacement of te secind shot. I.e, 40^2.sin120÷g = Vcos60.t => V^2-18V-1600 = 0 => V = 50/-32. So the required velocity is 50 m/s.
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