Question

Two stones are projected from the top of a cliff $h$ metres high, with the same speed $u$ so as to hit the ground at the same spot. If one of the stones is projected horizontally and the other is projected at an angle $\theta$ to the horizontal, then $\tan \theta$ equals :

• A. $\sqrt{\frac{2 u}{g h}}$
• B. $2 g \sqrt{\frac{u}{h}}$
• C. $2 h \sqrt{\frac{u}{g}}$
• D. $u \sqrt{\frac{2}{g h}}$

Answer 1

Answer: (D)

When stone is projected horizontally, then
$ R=u \cos 0 \times t$
and $h=u \sin 0 \times t+\frac{1}{2} g t^{2}$
$\Rightarrow R=u t$ and $h=\frac{1}{2} g t^{2}$
$\Rightarrow R=u t$ and $t=\sqrt{\frac{2 h}{g}}$
$\Rightarrow R=u \sqrt{\frac{2 h}{g}} \ldots$ (i)
When stone is projected at an angle of $\theta$ to the horizontal, then
$R=u \cos \theta \times t \dots$(ii)
and $ h=-u \sin \theta \times t+\frac{1}{2} g t^{2} \dots$(iii)
Using Eq. (i) in Eq. (ii), we get
$ u \sqrt{\frac{2 h}{g}}=u \cos \theta \times t$
$\Rightarrow t=\frac{1}{\cos \theta} \sqrt{\frac{2 h}{g}} \ldots$ (iv)
Using Eq. (iv) in Eq. (iii), we get
$ h=\frac{-u \sin \theta}{\cos \theta} \sqrt{\frac{2 h}{g}}+\frac{1}{2} g \frac{1}{\cos ^{2} \theta}\left(\frac{2 h}{g}\right) $
$\Rightarrow h=-u \tan \theta \sqrt{\frac{2 h}{g}}+\frac{h}{\cos ^{2} \theta} $
$\Rightarrow h\left[1-\frac{1}{\cos ^{2} \theta}\right]=-u \tan \theta \sqrt{\frac{2 h}{g}} $
$\Rightarrow h\left(-\tan ^{2} \theta\right)=-u \tan \theta \sqrt{\frac{2 h}{g}} $
$\Rightarrow \tan \theta=u \sqrt{\frac{2}{g h}}$