IIT JEE , AIEEE Forum - Physics , Mechanics - projectile cum rotation.....................challenging...........

jaysunantony

A particle of mass m is projected with a velocity u making 45 with
ground. The magnitude of angular momentum of the particle about the point of
projection, when it is at max. height h is :

a]zero

b]mu³/ 4.

2. g

c]mu³/

2 .g

d]m

[2gh³]

e]Both B and D

elessar_iitkgp

At max height, the velocity of the projectile is horizontal and equal to the X component of the initial velocity
v = ucos45 i= u/

2 i
r = R/2 i + Hj
Where R is the range and H is the height of the projectile.
L = m(v x r) = m(u/

2)H k = mu³/4

2g k
(As H = u²sin²45/2g = u²/4g )

jaysunantony

Thanx..............

but D option is again correct................So E is the correect one.............

And I got what I needed................thnx agn........

elessar_iitkgp

I forgot to see d. My mistake.
Anyways glad that I could be of help.

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