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A train is moving with a constant speed of 10 m/s in a circle of radius $\frac{16}{π}$ m. The plane of the circle lies in horizontal. x-y plane. At time t=0, train is at point $P (\frac{16}{π}, 0, 0)$ 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 $37^{\circ}$ with vertical z-axis.
Find the velocity of particle relative to train at the highest point of its trajectory.
$(- 6^i + 8^j) m s^{- 1}$
$(4^i + 10^j) m s^{- 1}$
$(- 3^i + 4^j) m s^{- 1}$
$(- 3^i + 5^j) m s^{- 1}$

A

(- 6^i + 8^j) m s^{- 1}

B

(4^i + 10^j) m s^{- 1}

C

(- 3^i + 4^j) m s^{- 1}

D

(- 3^i + 5^j) m s^{- 1}

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Solution

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The stone has three components of velocity.
The velocity of stone $v_{s} = v_{s / t r a i n} + v_{t r a i n}$
$v_{s / t r a i n} = - 10 c o s 37^{\circ}^i + 10 s i n 37^{\circ}^k m / s$ ; $v_{t r a i n} = 10^j$
$v_{s} = - 10 c o s 37^{\circ}^i + 10^j + 10 s i n 37^{\circ}^k m / s$
$_{s} = - 6^i + 10^j + 8^k$
At the highest point of projectile the vertical velocity will be zero
Velocity along z-direction is 8m/s, thus, time to reach max height is
$T = \frac{v_{z}}{g} = \frac{8}{10} = 0.8 s$
Velocities along x and y directions won't change as there is no aceleretion in these directions.At max height velocity of the particle is
$_{s} = - 6^i + 10^j$
Till this instant distance travelled by the train is $= 10 \times 0.8 = 8 m$
$\frac{8}{R} = \frac{8}{\frac{16}{π}} = \frac{π}{2} \Rightarrow$ angular displacement of the train is $90^{0}$
$_{t} = - 10^i_{p / t} =_{s} -_{t} = 4^i + 10^j$

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