Question

A ball is thrown from the ground to clear a wall 3 m high at a distance of 6 m and falls 18 m away from the wall. Find the angle of projection of ball.

• A. ${\tan }^{-1}\left(\frac{1}{3}\right)$
• B. ${\tan }^{-1}\left(\frac{2}{3}\right)$
• C. ${\tan }^{-1}\left(\frac{3}{2}\right)$
• D. ${\tan }^{-1}\left(\frac{3}{5}\right)$

Answer 1

Answer: (B)

${\tan }^{-1}\left(\frac{2}{3}\right)$

$R=6+18=24m=\frac{u^{2}sin2\theta }{g}\Rightarrow u^2=\frac{240}{2sin\theta cos\theta }=\frac{120}{sin\theta cos\theta }y=xtan\theta -\frac{g{x}^2}{2u^2{cos}^2\theta }\Rightarrow 3=6tan\theta -\frac{360}{2u^2{cos}^2\theta }\Rightarrow 3=6tan\theta -\frac{360}{240}tan\theta =\frac{9}{2}tan\theta \Rightarrow tan\theta =\frac{2}{3}$