Question

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

• A. $ta{n}^{-1}\left(\frac{2}{3}\right)$
• B. $ta{n}^{-1}\left(\frac{1}{4}\right)$
• C. $ta{n}^{-1}\left(\frac{3}{4}\right)$
• D. $ta{n}^{-1}\left(4\right)$

Answer 1

The correct option is A $ta{n}^{-1}\left(\frac{2}{3}\right)$

The range is R = 18 + 6 = 24m. We have
$y=xtan\theta (1-\frac{x}{R})\Rightarrow 3=6tan\theta (1-\frac{6}{24}).$
Thus $tan\theta =\frac{2}{3}\Rightarrow \theta =ta{n}^{-1}\left(\frac{2}{3}\right)$