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)/( 4) - alpha `
D. `(pi)/(2)`

Answer

A

Explanation

`H = ( u_(0)^(2) sin^(2) alpha)/( 2 g)`
Velocity at max height
`v_(1) = u_(0) cos alpha rArr v_(x) = u_(0) cos alpha `
When thrown vertically upward
`v_(2)^(2) = u_(0)^(2)` - 2gh
`rArr v_(2)^(2) = u_(0)^(2) - 2 xx g ( u_(0)^(2) sin^(2) alpha)/( 2 g)`
`rArr v_(2)^(2) = u_(0)^(2) cos^(2) alpha `
`rArr v_(2) = u_(0) cos alpha rArr v_(y) = u_(0) cos alpha `
`rArr tam theta = (v_(y))/( v_(x)) = 1 rArr theta = 45^(@)`