The velocity at the maximum height of a projectile is half its initial velocity of projection u Its range on the horizontal plane is

Question

The velocity at the maximum height of a projectile is half its initial velocity of projection u Its range on the horizontal plane is (A)  (√3u2/2g)  (B)  (u2/3g)  (C)  (3u2/2g)  (D)  (3u2/g)

Tardigrade Admin · Accepted Answer

There is only horizontal component of velocity at the highest point because at highest point vertical component of velocity is zero, body begins to come downwards from the highest point.

∴ u cos θ = \frac{u}{2} cos θ

= \frac{1}{2} = cos 6 0^{\circ}

\Rightarrow θ = 6 0^{\circ}

ange

R = \frac{2 u ^{2} s i n θ c o s θ}{g}

R = \frac{2 u ^{2} \times s i n 6 0 ^{\circ} c o s 6 0 ^{\circ}}{g}

R = \frac{3 u ^{2}}{2 g}

Q. The velocity at the maximum height of a projectile is half its initial velocity of projection $u$ Its range on the horizontal plane is

$\frac{3 u ^{2}}{2 g}$

$\frac{u ^{2}}{3 g}$

$\frac{3 u ^{2}}{2 g}$

$\frac{3 u ^{2}}{g}$

Q. The velocity at the maximum height of a projectile is half its initial velocity of projection u Its range on the horizontal plane is

2g3​u2​

3gu2​

2g3u2​

g3u2​

Q. The velocity at the maximum height of a projectile is half its initial velocity of projection $u$ Its range on the horizontal plane is

$\frac{3 u ^{2}}{2 g}$

$\frac{u ^{2}}{3 g}$

$\frac{3 u ^{2}}{2 g}$

$\frac{3 u ^{2}}{g}$