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) sin (2 xx 15^(@))) /(g) = 1.5 or (u^(2)) /(g) xx (1)/(2) = 1.5`
or `(u^(2))/(g) = 3 `km
Horizontal range for angle of projection `45^(@)` is
`R = (u^(2)) /(g) xx sin( 2xx 45^(@)) = (u^(2))/( g) sin90^(@) = (u^(2))/( g) = 3`km