Question

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

Answer 1

`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