A particle aimed at a target, projected with an angle `15^(0)` with the horizontal is short of the target by 10 m. If projected with an angle of 45° is away from the target by 15 m, then the angle of projection to hit the target is

Options

A. `(1)/(2)sin^(-1)((1)/(10))`
B. `(1)/(2)sin^(-1)((3)/(10))`
C. `(1)/(2)sin^(-1)((9)/(10))`
D. `(1)/(2)sin^(-1)((7)/(10))`

Answer

D

Explanation

According to question, `R_(1)=R+15`
and again with reference to question, `R_(2)=R+15`
Now as we know that, range `propsin2theta`,
So, `(R-10)/(R+15)=(sin2theta_(1))/(sin2theta_(2))=(sin30^@)/(sin90^@)=(1)/(2)`
`Rightarrow 2R-20=R+15RightarrowR=35`
For the range to be maximum `sin2theta=sin90^(@)=1`
So, `R_("max")=(u^2)/(g)`
As, `R_(1)=(u^(2)sin2theta)/(g)Rightarrow50=(u^2)/(g)`
where, `theta=` angle of projection and `35=(50gxxwsin2theta)/(g) [u^(2)=50g]`
`therefore sin2theta=(35)/(50)=(7)/(10) Rightarrow2theta=sin^(-1)[(7)/(10)]`
`Rightarrow theta=(1)/(2)sin^(-1)[(7)/(10)]`