A projectile is projected at an angles of `37^(@)` to the horizontal. Had it been projected at `53^(@)`, its time of flight on level ground would have been 1 sec more. At what angles should it be projected so that its range is 10 m less than either of the situations ?

Options

A. `26.5^(@)`
B. `30^(@)`
C. `18^(@)`
D. `15^(@)`

Answer

A

Explanation

`T_(1)=(2u sin 37^(@))/g`
`rArr T_(2)=(2u sin 53^(@))/g`
`(2u)/g (sin 53^(@)-sin37^(@))=1`
`24/10(4/5-3/5)=1`
u=25 m/s
`R=(25xx2xx3/5xx4/5)/10 = 60 m`
`R'=50 =(25 sin 2 theta)/10`
`4/5 =sin 2 theta`
`2 theta=53^(@)`