Question

A bdy is projected obliquely at an angle 30 degrees with the horizontal. The time when angular momentum is maximum in seconds

Answer 1

Answer:

Explanation:

the magnitude of angular momentum of the projectile about the point of projection is T = mvr

vx = ucos30 = u√3/2

vy = usin30 - gt = u/2 - 5t

v = √ ( u/2 - 5t)² + 3u²/4

x = ut√3/2 , y = ut/2 - 5t²/2

r = √(ut/2 - 5t²/2)² + 3u²t²/4

to  maximize M → dM/dt = M' = 0

let log M = logm + log v + log r

M'/M = v'/v + r'/r = 0

rv' + vr' = 0

- √{(ut/2 - 5t²/2)² + 3u²t²/4}{5t - u/2)/v∧3/2 +  {√ ( u/2 - 5t)² + 3u²/4 }{  2(ut/2 - 5t²/2)(u/2-5t) +  3u²t/2}/ r∧3/2 = 0