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Two seconds after projection a projectile is travelling in a direction inclined at $30^o$ to horizontal, after one more second it is travelling horizontally. What is the magnitude and direction of its velocity at initial point
$20 \sqrt 3 \,m/s, 30^o$
$20 \sqrt 3 \,m/s, 60^o$
$10 \sqrt 3 \,m/s, 30^o$
$10 \sqrt 3 \,m/s, 60^o$
Solution
$\mathrm{T}=\frac{2 \mathrm{u} \sin \theta}{\sigma}=6$ seconds
$4 \sin \theta=3 g \ldots \ldots(i)$
solve $(1)$ $\&(2)$ to get
$\tan \alpha=\frac{4 \sin \theta-g t}{4 \cos \theta}$
$\tan 30^{\circ}=\frac{4 \sin \theta-g \times 2}{4 \cos \theta}$
$\frac{1}{\sqrt{3}}=\frac{3 g-2 g}{4 \cos \theta}$
$4 \cos \theta=\sqrt{3} \mathrm{g}$ $…(2)$
Solve $(1)$ and $(2)$ to get
Similar Questions
Trajectory of particle in a projectile motion is given as $y=x-\frac{x^2}{80}$. Here, $x$ and $y$ are in metre. For this projectile motion match the following with $g=10\,m / s ^2$.
| $Column-I$ | $Column-II$ |
| $(A)$ Angle of projection | $(p)$ $20\,m$ |
| $(B)$ Angle of velocity with horizontal after $4\,s$ | $(q)$ $80\,m$ |
| $(C)$ Maximum height | $(r)$ $45^{\circ}$ |
| $(D)$ Horizontal range | $(s)$ $\tan ^{-1}\left(\frac{1}{2}\right)$ |
