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. (a)`1/(2)sin^(-1)1/(10)`
B. (b)`1/(2)sin^(-1)3/(10)`
C. (c)`1/(2)sin^(-1)9/(10)`
D. (d)`1/(2)sin^(-1)7/(10)`

Answer

(d)

Explanation

Given particle projected with angle `15^(0)` is short of the target by 10 m.
`R - 10 = U^(2)sin2(15^(0)/(g) = U^(2)/(2g))`
If projected with an angle of `45^(0)` is away from the target by 15 m.
`R + 15 = U^(2)sin2(45^(0)/(g) = U^(2)/(g))`
`rArr 2(R - 10) = R + 15`
`:. R = 35 m`
`rArr U^(2)/(g)= 35 + 15 = 50`
`sin2theta=35/(50)=7/(10)`
`:. theta = 1/2sin^(-1)(7/10)`