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Grade: 12th pass

                        

For an object projected from ground with speed u horizontal range is two times the maximum height attained by it. The horizontal range of object is

3 years ago

Answers : (1)

Lakshmi Vinod
29 Points
							R=2H....(Given)We know thatR=4Hcot(theta)2H=4Hcot(theta)1=2cot(theta)1=2×[1/tan(theta)]Hence tan(theta)=2Now by using Range FormulaR=u×usin2(theta)/g=u×u/g×(2tan(theta)/1+tan×tan(theta)=u×u(2×2)/g(1+2×2)=4u×u/5g
						
3 years ago
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