Question

the velocity at maximum height of a projectile is half its initial velocity (u). its range of horizontal plane is:
a) 3u²/g
b) 3/2 . u²/g
c) u/3g
d) √3/2 . u²/g

Answer 1

Answer:

• Range of projectile would be √3/2 . u²/g

Option (d) √3/2 . u²/g is correct.

Explanation:

Given that,

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

We need to find,

• Range of horizontal plane

So,

Let, angle of projection of projectile be θ

its initial velocity is given as u

and, Let its velocity at maximum height be v

then, By vector resolution there will be two components of initial velocity (u), vertical component as u sin θ and horizontal component  as u cos θ respectively.

so,

At maximum height vertical component would be zero because angle with horizontal plane would become zero [ u sin 0° = 0 ]

therefore, velocity at maximum height would become dependent on horizontal component of u

hence,

→ Velocity at max. height = horizontal component of initial velocity

→ v = u cos θ

(since, given that velocity at maximum height is half of initial velocity)

→ u / 2 = u cos θ

→ cos θ = 1/2

→ θ = 60°

Now,

Using formula for range (R) of horizontal plane

R = u² sin2θ / g

[ where R is range, u is initial velocity, θ is angle of projection, g is acceleration due to gravity ]

putting known values

→ R = u² sin2(60°) / g

→ R = u² sin 120° / g

→ R = u² (√3/2) / g

→ R = √3/2 . u² / g

Therefore,

• OPTION (d) √3/2 . u²/g is correct.

Answer 2

Answer:

✡ Given ✡

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

✡ To Find ✡

$\leadsto$ What its range of horizontal plane.

✡ Solution ✡

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

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

$\mapsto$ $\sf{V_x}$ = $\dfrac{u}{2}$

$\implies$ $\sf{u_y}$ = ${\sqrt{u²}-\dfrac{u²}{4}}$ = $\dfrac{√3 × u}{2}$

▶ tan$\theta$ = $\sqrt{60°}$

$\implies$ $\theta$ = 60°

➡ Range of projectile ,

$\implies$ $\dfrac{u²sin2\theta}{g}$

$\implies$ $\dfrac{u²sin120°}{g}$

$\implies$ $\dfrac{√3u²}{2g}$

Correct options is d.

Explanation:

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