Question

A projectile is projected with velocity of $25\, m / s$ 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 $g=10 \, m / s ^{2}$ ]

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

Answer 1

Answer: (D)

$R =\frac{ V ^{2}(2 \sin \theta \cos \theta)}{ g }$
$t =\frac{ V \sin \theta}{ g } \Rightarrow V =\frac{ gt }{\sin \theta}$
$\Rightarrow R =\frac{ g ^{2} t ^{2}}{\sin ^{2} \theta} \cdot \frac{2 \sin \theta \cos \theta}{ g }$
$\tan \theta=\frac{2 gt ^{2}}{ R }=\frac{20 t ^{2}}{ R }$
$\cot \theta=\frac{ R }{20 t ^{2}}$