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

What are the two angles of projection of a projectile projected with v

1.	What are the two angles of projection of a projectile projected with velocity of `30 m//s`, so that the horizontal range is ` 45 m`. Take, ` g= 10 m//s^(2)`.
Answer» Horizontal range `=(u ^(2) sin 2 theta)/g `45=(30^^(2) sin 2 theta)/(10)` `sin 2 theta =(45 xx 10)/(30 xx 30) =1/2 =sin 30^(@) ` or sin 15^(@)` ` 2 theta =30^(@) or 15^(@) or theta =15^(@) or 75%(@)`.

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

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