Question

A particle just clears a wall of height $b$ at a distance $a$ and strikes the ground at a distance $c$ from the point of projection. The angle of projection is:

• A. $ta{n}^{-1}\frac{b}{ac}$
• B. ${45}^0$
• C. $ta{n}^{-1}\frac{bc}{a(c-a)}$
• D. $ta{n}^{-1}\frac{ac}{a}$

Answer 1

Answer: (C)

$ta{n}^{-1}\frac{bc}{a(c-a)}$

$u\cos \theta t=a\Rightarrow t=\frac{a}{u\cos \theta }$

$b=u\sin \theta t-\frac{1}{2}g{t}^2$

$b=u\sin \theta \frac{a}{u\cos \theta }-\frac{1}{2}g\frac{a^2}{u^2{\cos }^2\theta }$ .............$\left(1\right)$

$c=u\cos \theta \times \frac{2u\sin \theta }{g}$

$\cos \theta =\frac{gc}{2u^2}$....................$\left(2\right)$

From $\left(1\right)$ and $\left(2\right)$,

$b=\frac{asin\theta }{cos\theta }-\frac{a^{2}sin\theta }{ccos\theta }$

Solving,

$\theta =ta{n}^{-1}\frac{bc}{a(c-a)}$