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

Answer 1

Answer: (D)

When stone is projected horizontally, then
$R=u\cos 0\times t$
and $h=u\sin 0\times t+\frac{1}{2}g{t}^2$
$\Rightarrow R=ut$ and $h=\frac{1}{2}g{t}^2$
$\Rightarrow R=ut$ and $t=\sqrt{\frac{2h}{g}}$
$\Rightarrow R=u\sqrt{\frac{2h}{g}}\dots$ (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
$u\sqrt{\frac{2h}{g}}=u\cos \theta \times t$
$\Rightarrow t=\frac{1}{\cos \theta }\sqrt{\frac{2h}{g}}\dots$ (iv)
Using Eq. (iv) in Eq. (iii), we get
$h=\frac{-u\sin \theta }{\cos \theta }\sqrt{\frac{2h}{g}}+\frac{1}{2}g\frac{1}{{\cos }^2\theta }\left(\frac{2h}{g}\right)$
$\Rightarrow h=-u\tan \theta \sqrt{\frac{2h}{g}}+\frac{h}{{\cos }^2\theta }$
$\Rightarrow h\left[1-\frac{1}{{\cos }^2\theta }\right]=-u\tan \theta \sqrt{\frac{2h}{g}}$
$\Rightarrow h\left(-{\tan }^2\theta \right)=-u\tan \theta \sqrt{\frac{2h}{g}}$
$\Rightarrow \tan \theta =u\sqrt{\frac{2}{gh}}$