A particle of mass `m` projected horizontally with velocity `u`. if it makes an angle `theta` with the horizontal after some time, then at that instant, its be Broglie wavelength is

Answer

4

Explanation

For a projectile horizontal component of velocity is constant.
`thereforev_(x)=u_(x),Vcostheta=u`
`therefore` de Broglie wavelenght , `lambda = (h)/(mv) = (hcos theta)/(mu)`