A missile is launched from the ground making 45 degree with the horizontal to hit a target at a horizontal distance of 300 km if it is required to hit a target at a horizontal distance of 675 km launched at same angle with horizontal find the percentage change in its velocity of projection.
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Expert's answer
2018-04-19T09:45:08-0400
The range is R=v^2/g sin2θ sin2θ=sin2(45)=1 R^'/R=(v^'/v)^2 (v^'/v)=√(R^'/R) (v^'/v)-1=√(R^'/R)-1=√(675/300)-1=0.5. Therefore, the percentage change in its velocity of projection is 50%.
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