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 $g =10 m / s ^{2}$ ]
$\frac{1}{2} \sin ^{-1}\left(\frac{5 t^{2}}{4 R}\right)$
$\frac{1}{2} \sin ^{-1}\left(\frac{4 R }{5 t ^{2}}\right)$
$\tan ^{-1}\left(\frac{4 t ^{2}}{5 R }\right)$
$\cot ^{-1}\left(\frac{ R }{20 t ^{2}}\right)$
A projectile is thrown in the upward direction making an angle of $60^o $ with the horizontal direction with a velocity of $147\ ms^{-1}$ . Then the time after which its inclination with the horizontal is $45^o $ , is ......... $\sec$
A particle of mass $100\,g$ is projected at time $t =0$ with a speed $20\,ms ^{-1}$ at an angle $45^{\circ}$ to the horizontal as given in the figure. The magnitude of the angular momentum of the particle about the starting point at time $t=2\,s$ is found to be $\sqrt{ K }\,kg\,m ^2 / s$. The value of $K$ is $............$ $\left(\right.$ Take $\left.g =10\,ms ^{-2}\right)$
A stone is thrown at an angle $\theta $ to the horizontal reaches a maximum height $H$. Then the time of flight of stone will be
Match the columns
Column $-I$ $R/H_{max}$ |
Column $-II$ Angle of projection $\theta $ |
$A.$ $1$ | $1.$ ${60^o}$ |
$B.$ $4$ | $2.$ ${30^o}$ |
$C.$ $4\sqrt 3$ | $3.$ ${45^o}$ |
$D.$ $\frac {4}{\sqrt 3}$ | $4.$ $tan^{-1}\,4\,=\,{76^o}$ |