$=\frac{20}{\sqrt{3}} \times \frac{1}{2}=\frac{10}{\sqrt{3}} \mathrm{m} / \mathrm{s}$
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| $Column-I$ | $Column-II$ |
| $(A)$ 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 | $(s)$ $\tan ^{-1}\left(\frac{1}{2}\right)$ |
Assertion $A$ : When a body is projected at an angle $45^{\circ}$, it's range is maximum.
Reason $R$ : For maximum range, the value of $\sin 2 \theta$ should be equal to one.
In the light of the above statements, choose the correct answer from the options given below :
