A ball with projection velocity `10 m//s` is to hit a target `6 m` away in the same horizontal line. How high above the target must the ball be aimed so that the ball will hit the target ? `(tan 18.5^(@) = 1//3, tan 71.5^(@) = 3)`

Explanation

To hit the target, range `= 6m`
`R = (u^(2))/(g)sin 2 theta`
`6 = ((10)^(2))/(10)sin 2 theta rArr sin 2 theta = (6)/(10) = (3)/(5) = sin 37^(@)`
`2 theta = 37^(@) rArr theta = 18.5^(@)`
Another angle of projection `= 90 - 18.5^(@) = 71.5^(@)`
If `theta = 18.5^(@), h = 6 tan theta = 6 tan (18.5^(@)) = 6 xx (1)/(3) = 2 m`
`theta = 71.5^(@), h = 6 tan theta = 6 tan (71.5^(@)) = 6 xx 3 = 18 m`