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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 | $(s)$ $\tan ^{-1}\left(\frac{1}{2}\right)$ |
$(A \rightarrow r, B \rightarrow r, C \rightarrow p, D \rightarrow q)$
$(A \rightarrow r, B \rightarrow s, C \rightarrow p, D \rightarrow q)$
$(A \rightarrow q, B \rightarrow r, C \rightarrow p, D \rightarrow s)$
$(A \rightarrow s, B \rightarrow r, C \rightarrow p, D \rightarrow q)$
Solution
(a)
Comparing with the standard equation of projectile,
$y=x \tan \theta-\frac{g x^2}{2 u^2 \cos \theta}$
$\begin{array}{ll}\text { We get } & \theta=45^{\circ} \\ \text { and } & u=20 \sqrt{2}\,m / s \end{array}$
Time period of this projectile is $4\,s$. Hence, after $4\,s$ velocity vector will again make $45^{\circ}$ with horizontal.
