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`        A cabin is moved up the inclined plane with constant acceleration g sin​​(theta). 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 particle perpendicular to inclined plane is same as range of particle with respect to the cabin parallel to palne then calculate value of cot(theta).`
4 years ago

Sudhanshu Saini
29 Points
```							8
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4 years ago
Sudhanshu Saini
29 Points
```							if u want the pic of soln u may give ur email id or smthing so that i can post it. :)
```
4 years ago
SRIJAN
11 Points
```							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 ATime 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^2For the ball to meet the incline/floor of cabin,  y = x * tan ASubstituting that we get:  t = (2 u2 Sec A) / gDisplacement along incline of the ball at t:  x sec A = (u1 - u2 tan A) tIn 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) * tEquating 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
```
one year ago
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### Course Features

• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions