A train is moving with a constant speed of 45km per hour. A passenger throws horizontally through the window from a height of 3 metres above the ground along the normal to the direction of motion of train with a velocity of 5 m/s. Neglecting air resistance determine the velocity with which the stone hits the ground.
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horizontal velocity of the train = 45 kmph = 45 * 5/18 = 12.5 m/s
horizontal velocity of the object wrt the train = 5 m/s
since the two above velocities are perpendicular to each other (in the horizontal plane), the resultant velocity (horizontal) of the object wrt stationary observer on the ground =
√ (5² + 12.5²) = 2.5 *√29 m/sec
When the object reaches the ground with initial vertical component of velocity =0, its vertical component = v
v² = u² + 2 g h = 0 + 2 * 10 * 3
hence v = 2 √15 m/sec
The resultant of the horizontal and vertical components is :
√ [ 181.25 + 60 ] = 15.52 m/sec
horizontal velocity of the object wrt the train = 5 m/s
since the two above velocities are perpendicular to each other (in the horizontal plane), the resultant velocity (horizontal) of the object wrt stationary observer on the ground =
√ (5² + 12.5²) = 2.5 *√29 m/sec
When the object reaches the ground with initial vertical component of velocity =0, its vertical component = v
v² = u² + 2 g h = 0 + 2 * 10 * 3
hence v = 2 √15 m/sec
The resultant of the horizontal and vertical components is :
√ [ 181.25 + 60 ] = 15.52 m/sec
meayush:
Thank u very much sir, but the answer also requires the direction of the velocity.
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