Question

Two stones are projected from the top of a cliff $h$ meters high, with the same speed $u$ so as 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 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

The correct option is D $u\sqrt{\frac{2}{gh}}$

When stone is projected horizontally then

and

$h=u\sin oxt+\frac{1}{2}g{t}^2$

$R=u\cos oxt$

$R=u\cos oxt$

$\Rightarrow R=ut$ and $h=\frac{1}{2}g{t}^2\Rightarrow R=ut$ and $t=\sqrt{\frac{2h}{g}}$

$\Rightarrow R=u\sqrt{\frac{2h}{g}}$ ------(1)

When stone is projected at an angle of $\theta$ to the horizontal, then

$R=u\cos \theta xt$ ------(2)

and

$h=-u\sin \theta xt+\frac{1}{2}g{t}^2$ --------(3)

Using equation (1) and (2) we get

$u\sqrt{\frac{2h}{g}}=u\cos \theta xt\Rightarrow t=\frac{1}{\cos \theta }\sqrt{\frac{2h}{g}}$ ---------(4)

Using equation (4) in equation (3) we get

$h=\frac{-u\sin \theta }{\cos \theta }\sqrt{\frac{2h}{g}}+\frac{1}{2}\frac{1}{{\cos }^2\theta }\left(\frac{2h}{g}\right)$

$\Rightarrow h=-u\tan \theta \sqrt{\frac{2h}{g}}+\frac{h}{{\cos }^2\theta }$

$\Rightarrow h(1-\frac{1}{{\cos }^2\theta })=-u\tan \theta \sqrt{\frac{2h}{g}}$

$\Rightarrow \frac{1}{2}(-{\tan }^2\theta )=-u\tan \theta \sqrt{\frac{2h}{g}}[\therefore 1-{\sec }^2\theta =-{\tan }^2\theta ]$

$\Rightarrow \tan \theta =\frac{2}{2h}$