The range of the projectile projected at an angle of 15^ with horizontal is 50 m . If the projectile is projected with same velocity at an angle of 45^ with horizontal, then its range will be :

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Correct answer: C. 100 m

Range of projectile R= v^2 2 g ( R (2 )) R_1 R_2 = (2 _1 ) (2 _2 ) = (2 15) (2 45) = 30^ 90^ 50 R _2 = 1 2 R _2=100 ~m

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