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)`.

Explanation

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%(@)`.