Q126 / 160 · TS EAMCET 2020 (11 Sep Shift 2) · Physics
A particle aimed at a target, projected with an angle $15^{\circ}$ with the horizontal is short of the target by $10 \mathrm{~m}$. If projected with an angle of $45^{\circ}$ is away from the target by $10 \mathrm{~m}$, then the angle of projection to hit the target is
- A$\frac{1}{2} \sin ^{-1}\left(\frac{1}{4}\right)$
- Correct answerB$\frac{1}{2} \sin ^{-1}\left(\frac{3}{4}\right)$
- C$\frac{1}{2} \sin ^{-1}\left(\frac{10}{4}\right)$
- D$\frac{1}{2} \sin ^{-1}\left(\frac{20}{4}\right)$
Solution
Let $d$ be the distance of target.
The horizontal range of projectile is
$R=\frac{u^2 \sin 2 \theta}{g}$
When $\theta=15^{\circ}$,
$R_1=\frac{u^2 \sin \left(2 \times 15^{\circ}\right)}{g}=\frac{u^2}{2 g}$
When $\theta=45^{\circ}$,
$R_2=\frac{u^2 \sin \left(2 \times 45^{\circ}\right)}{g}=\frac{u^2}{g}$
According to question,
$R_1=\frac{u^2}{2 g}=d-10$ $\ldots$ (i)
and $\quad R_2=\frac{u^2}{g}=d+10$ $\ldots$ (ii)
Dividing Eq. (i) by Eq. (ii), we get
$\frac{1}{2}=\frac{d-10}{d+10} \Rightarrow d=30 \mathrm{~m}$
From Eq. (ii), $\frac{u^2}{g}=30+10=40$ $\ldots$ (ii)
So, to hit the target at $30 \mathrm{~m}$,
$\left.30=\frac{u^2 \sin 2 \theta}{g} \Rightarrow 30=40 \sin 2 \theta \quad \text { [from Eq. (iii) }\right]$
$\Rightarrow \theta=\frac{1}{2} \sin ^{-1}\left(\frac{3}{4}\right)$
Hence, the correct answer is option: 'B'.