Question

Work out? The maximum range of projectile is 3000M.what would be its range when it is launched at an angle of 150 with the horizontal? Thanks

Answer 1

Horizontal range `R` of a projectile when projected at an angle `theta` with the horizontal with velocity `v_0` is given as

`R=(v_0^2sin\ 2theta)/g`
where `g` is acceleration due to gravity and `=9.81\ ms^-2`.

We know that maximum range is given for angle of projection `theta=45^@`. Therefore,

`R_max=(v_0^2)/g`

Inserting given values we get `v_0^2` as

`3000=(v_0^2)/9.81`
`=>v_0^2=(3000xx9.81)`

Now range for `theta=150^@`

`R_150=((3000xx9.81)\ sin\ (2xx150))/9.81`
`R_150=3000(-0.8660)`
`R_150=-2598\ m`
`-ve` sign shows that now the range is in the opposite direction.