MCQ
The ratio of the speed of a projectile at the point of projection to the speed at the top of its trajectory is $x$. The angle of projection with the horizontal is
- A$\sin ^{-1}(x)$
- B$\cos ^{-1}(x)$
- C$\sin ^{-1}(1 / x)$
- ✓$\cos ^{-1}(1 / x)$
✓
Answer
Correct option: D.
$\cos ^{-1}(1 / x)$
d
(d)
(d)
$\frac{u}{u \cos \theta}=x \operatorname{or} \cos \theta=\frac{1}{x} \Rightarrow \therefore \quad \theta=\cos ^{-1}\left(\frac{1}{x}\right)$
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