Question

A train is moving with a constant speed of $10m/s$ in a circle of radius $\frac{16}{\pi }$ m. The plane of the circle lies in horizontal. x-y plane. At time t=0 train is at point $P(16/\pi ,0,0)$ 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 ${37}^{\circ }$ 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 $g=10m/s^2,\sin {37}^{\circ }=\frac{3}{5}$

• A. $(-4.5m,8m,0),(0.3m,8m,3.2m)$
  
• B. $-2.5m,16m,0),(0.3m,16m,3.2m)$
  
• C. $(-4.5m,16m,0),(0.3m,8m,3.2m)$
  
• D. $(-2.5m,8m,0),(0.3m,16m,3.2m)$
  

Answer 1

The correct option is C $(-4.5m,16m,0),(0.3m,8m,3.2m)$

The point P is $(16/\pi ,0,0)$

As this point velocity of trains is $\overrightarrow{V_t}=10\hat{j}$

Relative Velocity of Stone w.r.t Trains is $\overrightarrow{V_{St}}=\overrightarrow{V_S}-\overrightarrow{V_t}=-6\hat{i}+8\hat{k}$

So the velocity of the stone is $-6i+10j+8k$

Now breaking motion into 3 separate direction (x,y&z)

Initial Velocities and acceleration are:

$u_{Sx}=-6;u_{Sy}=10;u_{Sz}=8a_{Sx}=0;a_{Sy}=0;a_{Sz}=-10$

Z-direction:

At halfway:

$v=0;v=u+at\therefore Timeofhalfflightis{t}_{1/2}=8/10=0.8s\therefore TotalTimeofflightist=0.6s$

Therefore max height is achieved after 0.8s and is:
$s=ut+\frac{1}{2}a{t}^2=8\times 0.8+(-10/2)\times .8\times .8=3.2$

After full flight the the stone will be in x-y plane $(z=0)$

X direction:

$x_i=\frac{16}{\pi }=5.1u_x=-6a_x=0Inhalfflight\left(maxheight\right)s=ut=-4.8\therefore x-coordinateis5.1-4.8=0.3Atfullflight(t=1.6s)s=ut=-9.6\therefore x-coordinateis5.1-9.6=-4.5$

Y direction:

$y_i=0u_y=10a_y=0Inhalfflight\left(maxheight\right)s=ut=8\therefore y-coordinateis8Atfullflight(t=1.6s)s=ut=16\therefore y-coordinateis16$