Question

Can anyone tell me the right solution of this.please send step by step explai nation

Answer 1

Answer:

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

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Answer 2

Answer:

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