MCQ
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 $(s)$ $\tan ^{-1}\left(\frac{1}{2}\right)$
  • $(A \rightarrow r, B \rightarrow r, C \rightarrow p, D \rightarrow q)$
  • B
    $(A \rightarrow r, B \rightarrow s, C \rightarrow p, D \rightarrow q)$
  • C
    $(A \rightarrow q, B \rightarrow r, C \rightarrow p, D \rightarrow s)$
  • D
    $(A \rightarrow s, B \rightarrow r, C \rightarrow p, D \rightarrow q)$

Answer

Correct option: A.
$(A \rightarrow r, B \rightarrow r, C \rightarrow p, D \rightarrow q)$
a
(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.

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