A ball of mass 1 kg is projected with a velocity of 20 √2 m / s from the origin of an xy co-ordinate axis system at an angle 45° with x-axis (horizontal). The angular momentum [In SI units] of the ball about the point of projection after 2 s of projection is [take .g=10 m / s 2] ( y-axis is taken as vertical)

Question

A ball of mass 1 kg is projected with a velocity of 20 √2 m / s from the origin of an xy co-ordinate axis system at an angle 45° with x-axis (horizontal). The angular momentum [In SI units] of the ball about the point of projection after 2 s of projection is [take .g=10 m / s 2] ( y-axis is taken as vertical) (A) -400 hatk (B) 200 hati (C) 300 hatj (D) -350 hatj

Tardigrade Admin · Accepted Answer

Time of flight T=(2 u sin θ/g)=(2(20 √2) (1/√2)/10)=4 second ⇒ After 2 second particle will be at maximum height of the projectile L=m v r perp r perp=H max =(u2 sin 2 θ/2 g)=20 m So L=(1)(20)(20)=400(- hatk)

$- 400 \hat{k}$

$200 \hat{i}$

$300 \hat{j}$

$- 350 \hat{j}$

Time of flight $T = \frac{2 u s i n θ}{g} = \frac{2 ( 20 2 ) \frac{1}{2}}{10} = 4$ second
$\Rightarrow$ After 2 second particle will be at maximum height of the projectile $L = m v r_{⊥}$
$r_{⊥} = H_{m a x} = \frac{u ^{2} s i n ^{2} θ}{2 g} = 20 m$
So $L = (1) (20) (20) = 400 (- \hat{k})$

−400k^

200i^

300j^​

−350j^​

Time of flight T=g2usinθ​=102(202​)2​1​​=4 second ⇒ After 2 second particle will be at maximum height of the projectile L=mvr⊥​ r⊥​=Hmax​=2gu2sin2θ​=20m So L=(1)(20)(20)=400(−k^)

$- 400 \hat{k}$

$200 \hat{i}$

$300 \hat{j}$

$- 350 \hat{j}$

Time of flight $T = \frac{2 u s i n θ}{g} = \frac{2 ( 20 2 ) \frac{1}{2}}{10} = 4$ second
$\Rightarrow$ After 2 second particle will be at maximum height of the projectile $L = m v r_{⊥}$
$r_{⊥} = H_{m a x} = \frac{u ^{2} s i n ^{2} θ}{2 g} = 20 m$
So $L = (1) (20) (20) = 400 (- \hat{k})$