A particle is projected with a velocity $u,$ at an angle $α,$ with the horizontal. At what time its vertical component of velocity becomes half of its net speed at the highest point will be

\frac{u}{2 g}

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\frac{\frac{u s i n α}{g} - u c o s α}{2 g}

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\frac{u}{2 g} (2 cos α - sin α)

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\frac{u}{2 g} (2 sin α - cos α)

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Solution

The correct option is D $\frac{u}{2 g} (2 sin α - cos α)$
net speed at highest point = ucos α

then

v = u -gt

ucos α/2 = usinα – gt

Then

$t = u s i n α / g - u c o s α / 2 g$
$\frac{u}{2 g} 2 s i n α - c o s α$

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A particle is projected with a velocity u, at an angle α, with the horizontal. At what time its vertical component of velocity becomes half of its net speed at the highest point will be

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A particle is projected with a velocity $u,$ at an angle $α,$ with the horizontal. At what time its vertical component of velocity becomes half of its net speed at the highest point will be

The correct option is D $\frac{u}{2 g} (2 sin α - cos α)$
net speed at highest point = ucos α

then

v = u -gt

ucos α/2 = usinα – gt

Then

$t = u s i n α / g - u c o s α / 2 g$
$\frac{u}{2 g} 2 s i n α - c o s α$