Question

The speed at the maximum height of a projectile
is half of its initial speed of projection (u). The
horizontal range of the projectile is
(1)√3u^2/2g
(2)√3u^2/g
(3)u^2/√3g
(4)2u^2/√3g​

Answer 1

Answer:

Answer in the attachment

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Answer 2

The velocity at maximum height of a projectile is half its initial velocity (u).

To Find

⇝ What its range of horizontal plane.

Solution

$\dfrac{Velocity\: at\: maximum\: height\:}{half\: of\: initial\: velocity\:}$

half of initial veloCity

Velocity at maximum height

$\dfrac{Velocity\: at\: maximum\: height\:}{horizontal\: component\:}$

horizontal component

Velocity at maximum height

$\mapsto↦ \sf{V_x}V x$

= $\dfrac{u}{2} 2u$

$\implies⟹ \sf{u_y}u y$

= ${\sqrt{u²}-\dfrac{u²}{4}} u² − 4u²$

= $\dfrac{√3 × u}{2} 2√3×u$

▶ $tan\thetaθ = \sqrt{60°}$

60°

$\implies⟹ \thetaθ = 60°$

Range of projectile ,

$\implies⟹ \dfrac{u²sin2\theta}{g} gu²sin2θ$

$\implies⟹ \dfrac{u²sin120°}{g} gu²sin120°$

$\implies⟹ \dfrac{√3u²}{2g} 2g√3u²$

Correct options is d.

Explanation:

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