Question

A cabin is moved up the inclined plane with constant acceleration gsin x . A particle is projected with some velocity with respect to the cabin in a direction perpendicular to the inclined plane. If maximum height attained by the particle to inclined plane is sane as range of particle with respect to the cabin parallel to plane then calculate values of cot x.

Answer 1

Let the angle of the incline = A.  (instead of x  , as given)

Let the cabin travel with an instantaneous velocity u1 up the incline at the instant the ball is thrown up at Origin (0,0) with a velocity u2 perpendicularly.

Deceleration of ball in direction perpendicular to incline:  - g cos A
Time to reach the max. height wrt incline = u2/(g cos A)

Maximum height wrt incline:   u2² /(2 g cos A)

For the range, we take the difference between the displacements of ball and origin when the ball meets the floor/incline.

Displacement of the ball:
           x = (u1 Cos A - u2 Sin A) t
           y = (u1 sin A + u2 Cos A) t - 1/2 * g t^2

For the ball to meet the incline/floor of cabin,  y = x * tan A
Substituting that we get:  t = (2 u2 Sec A) / g

Displacement along incline of the ball at t:  x sec A = (u1 - u2 tan A) t

In time t, the displacement of the cabin:   u1 t + 1/2 * g sin A t^2 

Range of ball relative to cabin = difference of both displ.
       =  (u2 tan A + 1/2 g t sin A) * t

Equating relative max height and relative range:
      u2^2 / (2g cos A) = (u2 tanA + g/2 SinA*2 u2 secA /g)*2 u2 secA /g
                                  = 4 tan A u^2 sec A / g

   =>  Cot A = 8