A particle of mass m is projected with a velocity v making an angle of 45° with the horizontal. The magnitude of the angular momentum of the projectile about the point of projection when the particle is at its maximum height h is

Question

A particle of mass m is projected with a velocity v making an angle of 45° with the horizontal. The magnitude of the angular momentum of the projectile about the point of projection when the particle is at its maximum height h is (A) zero (B) mv3 (4 √ 2 g ) (C)  mv3 (√ 2 g )  (D)  m √ 2 g h3

Tardigrade Admin · Accepted Answer

L = m (v /√ 2 ) r1 Here, r1 = h = (v2 sin 2 45 ° / 2g) = ( v2/4g) ∴ L = m ( (v/ √ 2) ) ( (v2/4g) ) = ( mv3/ 4 √ 2g)

Q. A particle of mass m is projected with a velocity v making an angle of 45 $^{\circ}$ with the horizontal. The magnitude of the angular momentum of the projectile about the point of projection when the particle is at its maximum height h is

zero

$m v^{3} (42 g)$

$m v^{3} (2 g)$

$m 2 g h^{3}$

$L = m \frac{v}{2} r_{1}$
Here, $r_{1} = h = \frac{v ^{2} s i n ^{2} 4 5 ^{\circ}}{2 g} = \frac{v ^{2}}{4 g}$
$∴ L = m (\frac{v}{2}) (\frac{v ^{2}}{4 g}) = \frac{m v ^{3}}{4 2 g}$

Q. A particle of mass m is projected with a velocity v making an angle of 45∘ with the horizontal. The magnitude of the angular momentum of the projectile about the point of projection when the particle is at its maximum height h is

zero

mv3(42​g)

mv3(2​g)

m2gh3​

L=m2​v​r1​ Here,r1​=h=2gv2sin245∘​=4gv2​ ∴L=m(2​v​)(4gv2​)=42g​mv3​

Q. A particle of mass m is projected with a velocity v making an angle of 45 $^{\circ}$ with the horizontal. The magnitude of the angular momentum of the projectile about the point of projection when the particle is at its maximum height h is

$m v^{3} (42 g)$

$m v^{3} (2 g)$

$m 2 g h^{3}$

$L = m \frac{v}{2} r_{1}$
Here, $r_{1} = h = \frac{v ^{2} s i n ^{2} 4 5 ^{\circ}}{2 g} = \frac{v ^{2}}{4 g}$
$∴ L = m (\frac{v}{2}) (\frac{v ^{2}}{4 g}) = \frac{m v ^{3}}{4 2 g}$