1.

The horizontal range of a projectile is times of its maximum height. Its angle of projection will be :-

Answer»

`45^(@)`
`60^(@)`
`90^(@)`
`30^(@)`

Solution :`R=(u^(2) SIN2THETA ) /(g)ANDH=(u^(2) sin ^(2)theta )/( 2g) `
but ,`R= 4 sqrt(3 ) H `
`(u^(2) sin2 theta) /(g)= 4 sqrt(3)((u^(2)sin ^(2)theta )/(2g))`
` sin 2 theta= 4 sqrt(3)(sin^(2)theta )/(2)`
`2 sintheta cos theta =(4sqrt(3) sin ^(2)theta )/(2)`
` (1)/(sqrt(3))-(sin theta )/(cos theta ) = tan theta `
or`theta = 30^(@)`


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