The angle of projection for a projectile to covermaximum range is :

1.	The angle of projection for a projectile to covermaximum range is :
Answer» `30^(@)` `60^(@)` `0^(@)` `45^(@)` SOLUTION :`R_(max")= (u^(2))/(g)` `(u^(2))/(g) SIN 2 theta = (u^(2))/(g) ` . `sin 2 theta = 1` `:. 2 theta = 90^(@)` `theta = 45^(@)`

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