Question

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^-1x
b.cos^-1x
c.sin^-1(1/x)
d.cos^-1(1/x)

Answer 1

Hope u like my process
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=> velocity at initial position of projectile = V m/s

=> at top of trajectory,

Velocity of vertical component = 0 m/s

Velocity at horizontal component =$v.\cos(\theta)$

So velocity at top of trajectory
= 0 +$v.\cos(\theta)$ =$v.\cos(\theta)$


So, the ratio is given as,

$= > \bf v : v \cos( \theta ) = x :1 \\ \\ = > \frac{1}{ \cos( \theta ) } = x \\ \\ = > \cos( \theta ) = \frac{1}{x} \\ \\ = > \bf \theta = \cos ^{ - 1} ( \frac{1}{x} )$
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Hence option d is the required answer.
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