Question

A ball is projected from ground with a speed of 30 m/s , find $\theta$ so that ball passes through point A. $(g=10m{s}^{-2})$

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

Answer 1

The correct option is B ${\tan }^{-1}\left(3\right)$

$y=x\tan \theta -\frac{g{x}^2}{2u^2{\cos }^2\theta }$

$40=30tan\theta -\frac{10\times \text{/}30\times \text{/}30}{2\times \text{/}30\times \text{/}30}{\sec }^2\theta$

$40=30\tan \theta -5{\sec }^2\theta$

$8=6\tan \theta -(1+{\tan }^2\theta )$

${\tan }^2\theta -6\tan \theta +9=0$

$\tan \theta =\frac{6\pm \sqrt{36-36}}{2}=\frac{6\pm 0}{2}=3$