MCQ
The horizontal range is four times the maximum height attained by a projectile. The angle of projection is .......... $^o$
- A$90$
- B$60$
- ✓$45$
- D$30$
$\frac{\mathrm{U}^{2} \times 2 \sin \theta \cos \theta}{\mathrm{g}}=4 \times \frac{\mathrm{U}^{2} \sin ^{2} \theta}{2 \mathrm{g}}$
$\tan \theta=1 \quad\left[\theta=45^{\circ}\right]$
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$Reason$ : Relative velocity of $P$ w.r.t. $Q$ is the ratio of velocity of $P$ and that of $Q$.
where $X = \frac{{{A^2}{B^{\frac{1}{2}}}}}{{{C^{\frac{1}{3}}}{D^3}}}$, will be
