Ratio of speed of a projectile at point of projection to speed at top of its trajectory is x . angle

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3-2.Motion in Plane

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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

$\sin ^{-1}(1 / x)$

D

$\cos ^{-1}(1 / x)$

Solution

(d)

$\frac{u}{u \cos \theta}=x \operatorname{or} \cos \theta=\frac{1}{x} \Rightarrow \therefore \quad \theta=\cos ^{-1}\left(\frac{1}{x}\right)$

Standard 11

Physics

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Column $-I$
Angle of projection Column $-II$

$A.$ $\theta \, = \,{45^o}$ $1.$ $\frac{{{K_h}}}{{{K_i}}} = \frac{1}{4}$

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$C.$ $\theta \, = \,{30^o}$ $3.$ $\frac{R}{H} = 4\sqrt 3 $

$D.$ $\theta \, = \,{\tan ^{ – 1}}\,4$ $4.$ $\frac{R}{H} = 4$

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