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What are two angles of projection of a projectile projected with velocity 30 m/s , so that the horizontal range is 45m . Take `g=10 m/s^2`.

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Horizontal range, `R=(u^2sin 2theta)/(g) or, sin 2theta =R*g//u^2`
`or, sin 2theta =(R*g)/(30^2)=(45xx10)/(30xx30) =1/2 =sin 30^@`
`therefore theta =15^@ and (90^@-15^@) =75^@`
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