A particle is projected from a point O with an initial speed of `30 ms^(-1)` to pass through a point which is `40 m` from O horizontally and `10 m` above O. There are two angles of projection for which this is possible. These angles are `alpha` and `beta`. The value of `|tan (alpha+beta)|` is

Answer

4

Explanation

`y=x tan theta-(9x^(2))/(2u^(2) cos^(2) theta)`
`10=40 tan theta-(10xx1600(1+tan^(2) theta))/(2xx900)`
`10=40 tan theta-80/9-80/9 tan^(2) theta`
`8/9 tan^(2) theta-4 tan theta+17/9=0`
`8 tan^(2) theta-36 tan theta+17=0`
`tan alpha+tan beta=36/8`
`tan alpha+tan beta=17/8`
`tan (alpha+beta)=(tan alpha+tan beta)/(1-tan alpha tan beta)`
`=(36//8)/(1-17/8)=(36//8)/(-9//8)=-4`
So `|tan(alpha+beta)|=|-4|=4`