Question

A stone is thrown from a point at a distance $a$ from a wall of height $b$. If it just clears the wall and angle of projection is $\alpha$, the maximum height attained by the stone is

• A. $\frac{a^2{\sec }^2\alpha }{4(a\sec \alpha -b)}$
• B. $\frac{a^2{\tan }^2\alpha }{4(a\tan \alpha -b)}$
• C. $\frac{a^2\tan \alpha }{4(a-b\sec \alpha )}$
• D. $\frac{a^2\tan \alpha }{4}$

Answer 1

The correct option is B $\frac{a^2{\tan }^2\alpha }{4(a\tan \alpha -b)}$

Equation of trajectory of the projectile is given by,
$y=x\tan \alpha -\frac{\left(g{x}^2\right)}{2u^2{\cos }^2\alpha }$
Height of wall, $y=b$ and distance between wall and point of projection, $x=a$, therefore
$b=a\tan \alpha -\frac{g{a}^2}{2u^2{\cos }^2\alpha }$
$\Rightarrow a\tan \alpha -b=\frac{g{a}^2}{2u^2{\cos }^2\alpha }$
therefore,
$\frac{u^2}{g}=\frac{a^2}{2{\cos }^2\alpha (a\tan \alpha -b)}$
$\text{Maximum height,}h=\frac{u^2{\sin }^2\alpha }{2g}$
$\Rightarrow h=\frac{a^2{\tan }^2\alpha }{4(a\tan \alpha -b)}$