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Question

# A train is moving with a constant speed of 10 m/s in a circle of radius 16π m. The plane of the circle lies in horizontal. x-y plane. At time t=0 train is at point P(16/π,0,0) and moving in counter-clockwise direction. At this instant a stone is thrown from the train with speed 10 m/s relative to train towards negative x-axis at an angle of 37∘ with vertical z-axis. The co-ordinates of points on the ground where it finally falls and that of the highest point of its trajectory.Take g=10m/s2,sin37∘=35

A
(4.5m,8m,0),(0.3m,8m,3.2m)
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B
2.5m,16m,0),(0.3m,16m,3.2m)
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C
(4.5m,16m,0),(0.3m,8m,3.2m)
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D
(2.5m,8m,0),(0.3m,16m,3.2m)
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Solution

## The correct option is C (−4.5m,16m,0),(0.3m,8m,3.2m)The point P is (16/π,0,0)As this point velocity of trains is →Vt=10^jRelative Velocity of Stone w.r.t Trains is →VSt=→VS−→Vt=−6^i+8^kSo the velocity of the stone is −6i+10j+8kNow breaking motion into 3 separate direction (x,y&z)Initial Velocities and acceleration are:uSx=−6;uSy=10;uSz=8aSx=0;aSy=0;aSz=−10Z-direction:At halfway: v=0;v=u+at∴Timeofhalfflightist1/2=8/10=0.8s∴TotalTimeofflightist=0.6sTherefore max height is achieved after 0.8s and is:s=ut+12at2=8×0.8+(−10/2)×.8×.8=3.2After full flight the the stone will be in x-y plane (z=0)X direction:xi=16π=5.1ux=−6ax=0Inhalfflight(maxheight)s=ut=−4.8∴x−coordinateis5.1−4.8=0.3Atfullflight(t=1.6s)s=ut=−9.6∴x−coordinateis5.1−9.6=−4.5Y direction:yi=0uy=10ay=0Inhalfflight(maxheight)s=ut=8∴y−coordinateis8Atfullflight(t=1.6s)s=ut=16∴y−coordinateis16

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