Banner 0
Banner 1
Banner 2
Banner 3
Banner 4
Banner 5
Banner 6
Banner 7
Banner 8
Banner 9

Q.

A particle of mass m is projected with a velocity v making an angle of 450 with the horizontal. The magnitude of the angular momentum of the projectile about the point of projection when the particle is at maximum height h is mv3(xg). Find the value of x.

see full answer

High-Paying Jobs That Even AI Can’t Replace — Through JEE/NEET

🎯 Hear from the experts why preparing for JEE/NEET today sets you up for future-proof, high-income careers tomorrow.
An Intiative by Sri Chaitanya

answer is 0032.00.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

JEE

NEET

NEET

Foundation JEE

Foundation JEE

Foundation NEET

Foundation NEET

CBSE

CBSE

Detailed Solution

At the highest point, velocity =vcos450=v2 in horizontal direction ​∴ Momentum =mv2​L= Angular momentum = Momentum × Perpendicular distance ​⇒L=mv2×h​ Here h=v2sin24502g=v24g​∴L=mv2v24 g=mv342 g=mv332 g
Watch 3-min video & get full concept clarity

courses

No courses found

score_test_img

Get Expert Academic Guidance – Connect with a Counselor Today!

whats app icon