A body is projected with velocity u making an angle - with the horizontal- Its velocity when it is perpendicular to the initial velocity vector u is-

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The Equation for the Path of Projectile

A body is pro...

Question

A body is projected with velocity $u$ making an angle $α$ with the horizontal. Its velocity when it is perpendicular to the initial velocity vector $u$ is:

u sin α

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u cot α

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u tan α

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u cos α

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Solution

The correct option is B $u cot α$
At time $t = 0$ (when body is projected),

Horizontal velocity, $u_{x} = u c o s α$

Vertical velocity, $u_{y} = u s i n α$

Velocity vector, $\to u = u c o s α^i + u s i n α^j$

At any time $t$ ;

Horizontal velocity, $v_{x} = u c o s α$

Vertical velocity, $v_{y} = u s i n α - g t$

Velocity vector, $\to v = u c o s α^i + (u s i n α - g t^) j$

When two vectors are perpendicular, their dot product is zero.

$\to u . \to v = 0$

This implies $({u c o s α)}^{2} + {(u s i n α)}^{2} - (u s i n α \times g t)$ = $0$

$t = \frac{u}{g s i n α}$

$∴ v_{y} = u s i n α - g (\frac{u}{g s i n α}) = u (s i n α - c o s e c α)$
Velocity magnitude, $| v | = \sqrt{({(u c o s α)}^{2} + {(u (s i n α - c o s e c α))}^{2}} = u c o t α$

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A body is projected with velocity u making an angle α with the horizontal. Its velocity when it is perpendicular to the initial velocity vector u is:

A body is projected with velocity $u$ making an angle $α$ with the horizontal. Its velocity when it is perpendicular to the initial velocity vector $u$ is: