The range of a particle when launched at an angle of `15^(@)` with the horizontal is 1.5 km. what is the range of the projectile when launched at an angle of `45^(@)` to the horizontal ?

Explanation

`R=(u^(2)sin2xx15)/(g)=1.5` or `(u^(2))'/(g)xx(1)/(2)=1.5` or `(u^(2))/(g)=3km`
Horizontal range for angle of projection `45^(@)` is `R=(u^(2))/(g)xxsin2xx45^(@)=(u^(2))/(g) sin 90^(@)=(u^(2))/(g)=3km`