Question

Galileo writes that for angle of projection of a projectile at angle 45+theta and 45-theta the horizontal ranges described by the projectile are in the ratio of

Answer 1

As we know that,

R = u² Sin 2Ø/g

For first case when Ø = (45° - Ø).

R = u² Sin 2(45°-Ø)/g

R = u² Sin (90° - 2Ø)/g

R = u² Cos 2Ø/g ------(1)

For second case when Ø = (45° + Ø)

R' = u² Sin (90° + 2Ø)/g

R' = u² Cos 2Ø/g

Hence, Ratio

R/R' = u² Cos 2Ø/g × g/u² Cos 2Ø

R/R' = 1/1

R:R' = 1:1 Answer.......