Question

A projectile is projected with velocity of $25\text{m/s}$ at an angle $\theta$ with the horizontal. After $t$ seconds its inclination with horizontal becomes zero. If $R$ represents horizontal range of the projectile, the value of $\theta$ will be
[use $g=10{\text{m/s}}^2$]

• A. ${\tan }^{-1}\left[\frac{4t^2}{5R}\right]$
• B. $\frac{1}{2}{\sin }^{-1}\left[\frac{5t^2}{4R}\right]$
• C. ${\cot }^{-1}\left[\frac{R}{20t^2}\right]$
• D. $\frac{1}{2}{\sin }^{-1}\left[\frac{4R}{5t^2}\right]$

Answer 1

The correct option is C ${\cot }^{-1}\left[\frac{R}{20t^2}\right]$


$t=\frac{25\sin \theta }{g}$
and, $R=\frac{(25{)}^2\left(2\sin \theta \cos \theta \right)}{g}$
$\Rightarrow R=\frac{25\times 25\times 2}{g}\times \frac{gt}{25}\times \cos \theta$
$\Rightarrow R=50t\cos \theta$
$\therefore \tan \theta =\frac{gt}{25}\times \frac{50t}{R}$
$=\frac{20t^2}{R}$
$\Rightarrow \theta ={\cot }^{-1}\left[\frac{R}{20t^2}\right]$