A particle of mass m is projected from the ground with an initial speed `u_0` at an angle `alpha` with the horizontal. At the highest point of its trajectory, it makes a completely inelastic collision with another identical particle, which was thrown vertically upward from the ground with the same initial speed `u_0.` The angle that the composite system makes with the horizontal immediately after the collision is

Options

A. `(pi)/(4)`
B. `(pi)/(4)+alpha`
C. `(pi)/(2)-alpha`
D. `(pi)/(2)`

Answer

A

Explanation

`H=(u_(0)^(2)sin^(2)alpha)/(2g)`
`v^(2)-u_(0)^(2)=(2(-g)u_(0)^(2)sin^(2)alpha)/(2g)`
`v^(2)=u_(0)^(2)cos^(2)alpha`